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NEW 



PLANE AND SPHERICAL 



TRIGONOMETRY 



BY 



WEBSTER WELLS, S.B. 

PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS 
INSTITUTE OF TECHNOLOGY 




LEACH, SHEWELL, AND SANBORN 

BOSTON NEW YORK CHICAGO • 






■VH3 



CLcrpM 



Copyright, 1896, 
By WEBSTER WELLS, 






/v, ' c i «v ° 



S. Cushing & Co. — Berwick & Sinitii 
Norwood Mass. U.S.A. 



PREFACE. 



In revising his Plane and Spherical Trigonometry, the author has 
effected many important improvements. The attention of teachers is 
specially invited to the following features of the new work : 

1. The proofs of the functions of 0°, 90°, 180°, and 270° ; §§ 22 to 25. 

2. The proofs of the functions of 120°, 135°, etc. ; § 27. 

3. The method of finding the values of the remaining functions of an 
angle when the value of any one is given ; § 28. 

4. The proofs of the functions of (— A), and (90° + A), in terms of 
those of A; §§29, 30. 

5. The method of solution in the examples of §§ 34 and 35. 

6. The general demonstration of the formulae for sin (x -j- y) and 
cos (x + y) ; § 42. 

7. The discussion of the line values of the functions, and their appli- 
cation in tracing the changes in the six principal functions of an angle 
as the angle increases from 0° to 360°; §§ 60, 61. 

8. The- discussion of trigonometric equations in § 62. 

9. The solution of right triangles by Natural Functions ; see Ex. 1, 
page 54. 

10. The discussion of the ambiguous case in the solution of oblique 
triangles; §§ 117 to 120. 

11. The proof of the formulae for the values of x in the cubic equation 
x i -ax-b = 0' i § 126. 

12. The geometrical proof of the important theorems of § 133. 

13. The demonstration of the formulae for right spherical triangles 
.before those for oblique spherical triangles ; see Chapters XL and XII. 

14. The reduction of the number of cases in the complete demonstra- 
tion of the fundamental theorems for spherical right triangles, to three, 
b}' application of the theorems of § 133 ; see § 136. 



:v 



PREFACE. 



15. The solution of Quadrantal and Isosceles Spherical triangles; 
§§ 149. 150. 

16. The discussion of the ambiguous cases in the solution of oblique 
spherical triangles; §§ 165, 166; especially the rules given on pages 10S 
and 111 for determining the number of solutions. 

At the end of Chapter XII. will be found a collection of formulae in 
form for convenient reference. 

The revised work contains a much greater number of examples than 
the old ; they have been selected with great care, and are with few excep- 
tions new. 

The results have been worked out by aid of the author's neAv Six Place 
Logarithmic Tables, which contain also a Table of Natural Functions, and 
an Auxiliary Table for Small Angles. The Trigonometry can be obtained 
either with or without the Tables. 

WEBSTER WELLS. 
Massachusetts Institute of Technology, 1896. 



CONTENTS. 



PLANE TRIGONOMETRY. 

Pa(;k 

I. Trigonometric Functions of Acute Angles 1 

II. Trigonometric Functions of Angles in General (5 

III. General Formulae 19 

IV. Miscellaneous Theorems 29 

V. Logarithms 41 

Properties of Logarithms 43 

Applications 47 

Exponential Equations 51 

VI. Solution of Right Triangles . . 54 

Formulae for the Area of a Right Triangle ...... 60 

VII. General Properties of Triangles . . 62 

Formulae for the Area of an Oblique Triangle 65 

VIII. Solution of Oblique Triangles 67 

Area of an Oblique Triangle 74 

IX. Cubic Equations 77 

SPHERICAL TRIGONOMETRY. 

X. Geometrical Principles , . SO 

XT. Right Spherical Triangles 83 

Solution of Right Spherical Triangles 88 

XII. Oblique Spherical Triangles 90 

General Properties of Spherical Triangles 9G 

Napier's Analogies 101 

Solution of Oblique Spherical Triangles 102 

Applications Ill 



Formulae; Plane Trigonometry 114 

Spherical Trigonometry 116 

Answers 119 

v 



PLANE TRIGONOMETRY. 



£<«< 



I. TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES. 

1. Trigonometry treats of the properties and measurement of angles 
and triangles. 

In Plane Trigonometry we consider plane figures only. 

2. Definitions of the Trigonometric Functions of Acute Angles. 

Let BAC be any acute angle. 




From any point in either side, as B, draw a perpendicular to the other 
side, forming the right triangle ABC. 

We then have the following definitions, applicable to either of the 
acute angles A or B : 

In any right triangle, 

The sine of either acute angle is the ratio of the opposite side to the 
hypotenuse. 

The cosine is the ratio of the adjacent side to the hypotenuse. 
The tangent ts the ratio of the opposite side to the adjacent side. 
The cotangent is the ratio of the adjacent side to the opposite side. 
The secant is the ratio of the hypotenuse to the adjacent side. 
The cosecant is the ratio of the hypotenuse to the opposite side. 

We also have the following definitions : 
The versed sine of an angle is 1 minus the cosine of the angle. 
The coversed sine is 1 minus the sine. 

The eight ratios defined above are called the Trigonometric Functions 
of the angle. 

1 



2 PLANE TRIGONOMETRY. 

Representing the sides BC, CA, and AB by a, b, and c, respectively, 
and employing the usual abbreviations, we have : 

A a 

sm A = — 
c 

cos ^4 = — 
c 

sin B=- 
c 

n a 

cos B = -- 

c 

3. It is important to observe that the values of the trigonometric 
functions depend solely on the magnitude of the angle, and are entirely 
independent of the lengths of the sides of the right triangle which con- 
tains it. 



tan .4 = -. 
b 


b 


vers A = l 


b 
c 


cot A = — 
a 


A C 

a 


covers J. = 1 


a 
c 


tan B = — 

a 


Q 

sec B = -' 
a 


vers B — l 


a 


cot. 5=-. 
b 


esc B = -> 

b 


covers B = 1 


b 

c 




For let B and B[ be any two points in the side AD of the angle DAE, 
and draw BC and B'C perpendicular to AE. 
Then by the definition of § 2, we have 

. A BC A ■ A B'C 

sn\A = , -and sml = , — ; — 

AB AB' 

But since, the right triangles ABC and AB'C are similar, their homol- 
ogous sides are proportional ; whence, 

BC^B'C 
AB AB'' 

Thus the two values obtained for sin A are equal. 
4. We have from § 2, 

a a 'A • & n b j n a 

sin .4 = -, cos^L = -, smJB=3-, and cosJ3=a — 
c c c c 

Whence, a = c sin i~c cos B, and 6 = c sin B = c cos A. 



TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES. 



That is, in any right triangle, either side about the right angle is equal to 
the hypotenuse multiplied by the sine of the opposite angle, or by the cosine 
of the adjacent angle. 

Again, 



tan ^4 = -, cotA = -, tan B — -, and cot !?:=-• 
baa b 



Whence, a = b tan A = b cot B, and b = a tan B = a cot A. 

That is, in any right triangle, either side about the right angle is equal 
to the tangent of the opposite angle, or the cotangent of the adjacent angle, 
multiplied by the other side. 



sin A = - = cos B. 
tan A — — — cot B. 



sec A 



esc B. 



vers A = l = covers B. 



5. We have from § 2, 
a 
c 
a 
b c 

As B is the complement of A, these results may be stated as follows : 
The sine, tangent, secant, and versed sine of any acute angle are respec- 
tively the cosine, cotangent, cosecant, and coversed sine of the complement of 
the angle. 

6. To Find the Values of the Other Seven Functions of an Acute Angle, 
when the Value of Any One is Given. 

1. Given esc A = 3; find the values of the remaining functions of A. 

B 




We may write the equation esc A 



Since the cosecant is the hypotenuse divided by the opposite side, we 
may regard A as one of the acute angles of the right triangle ABC, in 
which the hypotenuse AB == 3, and the opposite side BC = 1. 

Whence by Geometry, AC = V AB 2 - BC 2 = V9^T = V8 = 2 V2. 
Then by the definitions of § 2, 



sin^l 



cos A 



1 
3* 
2V2 



tan A = 



:V2 



sec A s= 



2V2 



cot A = 2 V2. 



vers A = l 



covers A = 1 = - 



2V2 
3 

3 3 



PLANE TRIGONOMETRY. 



2. Given vers A = - ; find the value of cot A. 
5 




Since vers A = l — cos A, we have eos A = 1 — vers A = l = -• 

o 5 

Then, in the right triangle ABC, we take the adjacent side AC = 3, 
and the hypotenuse AB = 5. 

Whence, BC = ^ AB 2 - AC 2 = V25 - 9 = Vl6 = 4. 

3 



Then by definition, 



cot A 





EXAMPLES. 




In each of the following, find the 


values of the remaining functions : 


3. sinA = ?- 
5 


5. cotA = — • 
24 


» A 3V3 
7. cos .4=— ^—- 
14 


9. sec A=.t. 


4. vers A = — • 


6. esc A = 7. 


Q ^ 2 
8. covers A =—• 


10. tan -4=-. 
a- 



21 
14. Given cos A= — ; find esc A. 



11. Given cot A = -; find sin .A 

12. Given cscA = — ; find cos A. 15. Given tan A: 

40 

13. Given sec A = 5; find cot A. 
7. Functions of 45°. 



4V2 



find sec A 



16. Given sin A= -: find tan A. 

x 




Let ABC be an isosceles right triangle, AC and BC being each equal 



) 1. 



Then ZA = 45°, and AB = V^LC 2 + .BC 2 = Vl + 1 = V2. 



TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES. 

AYhence by definition, 
sin 45 c 



1 =lV2. 



V2 2 

cos 45° =— =1 a/2. 

V2 2 

tan 45° = 1. 
cot 45° = 1. 



sec45°=V2. 

csc45°=V2. 

vers 45° = 1 -\ V2 = 2 ~ V ^ - 
covers 45° = 1 - ~ V2 = 2 ~ ^ 



8. Functions of 30° and 60°. 




Let ABD be an equilateral triangle having each side equal to 2. 

Draw AC perpendicular to BD. 

Then by Geo metry, BC= ±BD = 1, and Z BAC = \ Z BAD = 30 c 

Also, AC = ^AB 2 -BC 2 = Vi^l = V3. 

Then by definition, in the right triangle ABC, 



sin 30° = 
cos 30° = 



1 
V3 



= cos 60°. 
= sin 60°. 



tan 30° = — = 1 V3 = cot 60 c 

V3 3 



sec 30°=— =?V3 = esc 60°. 
V3 3 

esc 30° =2 = sec 60°. 

vers 30° = 1 - ^? = covers 60 c 



cot 30° = V3 



= tan 60°. 



covers 30°= 1- 



2 

1 = 1 

2 2 



vers 60°. 



6 PLANE TRIGONOMETRY. 

II. TRIGONOMETRIC FUNCTIONS OF ANGLES IN 

GENERAL. 

9. In Geometry, we are, as a rule, concerned with angles which arc 
less than two right angles; but in Trigonometry it is convenient to 
consider them as unrestricted in magnitude. 




A'" 



Let AA" and A'A'" be a pair of perpendicular diameters of the circle 
AA". 

Suppose a radius OB to start from the position OA, and revolve 
about the point as a pivot, in a direction contrary to the motion of the 
hands of a clock. 

When OB coincides with OA', it has generated an angle of 90°; when 
it coincides with OA", of 180°; with OA'", of 270°; with OA. its first 
position, of 360° ; with OA' again, of 450° ; and so on. 

We thus see that a significance may be attached to a positive angle 
of any number of degrees. 

10. The interpretation of an angle as measuring the amount of rota- 
tion of a moving radius, enables us to distinguish between positive and 
negative angles. 

Thus, if a positive angle indicates revolution from the position OA in 
a direction contrary to the motion of the hands of a clock, a negative 
angle may be taken as indicating revolution from the position OA in the 
same direction as the motion of the hands of a clock. 

Thus, if the radius OB' starts from the position OA, and revolves 
about the point O as a pivot in the same direction as the motion of the 
hands of a clock, when it coincides with OA'" it has generated an angle 
of -90°; when it coincides with OA", of -180°; with OA', of -270°; 
and so on. 

We may then conceive of negative angles of any number of degrees. 

It is immaterial which direction we consider the positive direction of 
rotation ; but having at the outset adopted a certain direction as positive, 
our subsequent operations must be in accordance. 



TRIGONOMETRIC FUNCTIONS OF ANGLES IN GENERAL. 7 

11. The fixed line OA from which the rotation is supposed to com- 
mence, is called the initial line, and the rotating radius in its final position 
is called the terminal line. 

12. In designating an angle, we shall always write first the letter at 
the extremity of the initial line. 

Thus, in designating the angle formed by the lines OA and OB, if 
we regard OA as the initial line, we should call it A OB-, and if we regard 
OB as the initial line, we should call it BOA. 

There are always two angles less than 360° in absolute value, one posi- 
tive and the other negative, formed by the same initial and terminal line. 

Thus, there are formed by OA and OB' the positive angle AOB' 
between 270° and 360°, and the negative angle AOB' between 0° and 
- 90°. 

We shall distinguish between such angles by referring to them as 
"the positive angle AOB'" and "the negative angle AOB'," respectively. 

13. It is evident that the terminal lines of any two angles which differ 
by a multiple of 360° are coincident. 

Thus, the angles 30°, 390°, — 330°, etc., have the same terminal line. 

14. Rectangular Co-ordinates. 



x : 



P 2 (-b,a) 
a 



N 
a 

P 3 (-b,-a) 



P x (h,a) 



P* (b,-a) 



Let Pj be any point in the plane of the lines XX' and YY' inter- 
secting at right angles at O, and draw P X M perpendicular to XX'. 

Then OM and P X M are called the rectangular co-ordinates of P x ; 031 
is called the abscissa, and P X M the ordinate. 

The lines of reference, XX' and YY', are called the axis of X and the 
axis of Y, respectively, and O is called the origin. 

It is customary to express the fact that the abscissa of a point is b, 
and its ordinate a, by saying that for the point in question x = b and 
y = a; or, more concisely, we may refer to the point as "the point 
(b, a)," where the first term in the parenthesis is understood to be the 
abscissa, and the second term the ordinate. 



8 PLANE TRIGONOMETRY. 

15. If, in the figure of § 14, OM= ON=b, and P Y P 4 and P 2 P S are 
drawn perpendicular to XX' so that P X M '= P 2 N '= P 3 X = P 4 M= a, the 
points Pj, P 2 , P 3 , and P 4 will have the same co-ordinates, (b, a). 

To avoid this ambiguity, abscissas measured to the right of are 
considered positive, and to the left, negative; and ordinates measured above 
XX' are considered positive, and below, negative. 

Then the co-ordinates of the points will be as follows : 

P 1? (b, a) ; P 2 , (- 6, a) ; P 3; (- 6, - a) ; P 4? (6, - a). 

16. If a point lies upon XX', its ordinate is zero ; and if it lies upon 
YV, its abscissa is zero. 

17. General Definitions of the Functions. 

AVe will now give general definitions of the trigonometric functions, 
applicable to any angle whatever. 

Take the initial line of the angle as the positive direction of the axis 
of X, the vertex being the origin. 

From any point in the terminal line, drop a perpendicular to the axis 
of X 

Find the co-ordinates of this point ; then, 

The sine of the angle is the ratio of the ordinate of the point to its 
distance from the origin. 

The cosine is the ratio of the abscissa to the distance. 

The tangent is the ratio of the ordinate to the abscissa. 

The cotangent is the ratio of the abscissa to the ordinate. 

The secant is the ratio of the distance to the abscissa. 

The cosecant is the ratio of the distance to the ordinate. 

18. We will now apply the definitions of § 17 to finding the functions 
of the angles XOP 2 , XOP 6 , and XOP 4 in the following figures : 



P 2 (-b,a) 




Pi(b,-a) 



Let P 2 , P 3 , and P 4 be any points on the terminal lines OP,, OP s , and 
OP 4 , and draw P 2 M, P 3 M, and P A M perpendicular to XX'. 

Let P 2 M= P 3 M= PJf= a, 031= b, and OP, = OP, = OP 4 = c. 



TRIGONOMETRIC FUNCTIONS OF ANGLES IN GENERAL. 9 

Then the co-ordinates of P 2 are (— b, a); of P 3 , (— b, —a); of P 4 , 
(6, - a). 



Whence by definition, 






sin XOP 2 = -• 

c 


. —a a 
sm A OP, = = 

c c 


sin XOP 4 = — - = 

c 


a 

c 


cos XOP 2 *= — =--• 

c c 


cos XOP 3 = — = --. 

3 c c 


cos XOP 4 = h '• 
c 




tan XOP 2 = — = - -• 
-6 b 


taiiXOP 3 = — =-• 

3 -6 5 


tan XOP 4 = ^" = 


a 
~b' 


cot XOP 2 = — = _-. 

a a 


cot XOP 3 = — ==-. 

— a a 


cot XOP 4 = — = 
— a 


b 
a 


sec XOP 2 = — = - £ 

— b b 


sec XOP 3 = — = - -• 

— b b 


sec XOPt = -. 
b 




esc XOP 2 = -- 

a 


CSC XOP3 = — = - -• 

— a a 


esc XOP 4 = -^- = 

— a 


c 
a 



Note 1. The definitions of § 17 are seen to include those of § 2. 

The definitions of the versed sine and coversed sine, given in § 2, are sufficiently 
general to apply to any angle whatever. 

Note 2. In all the figures of the present chapter, the small letters will be under- 
stood as denoting the lengths of the lines to which they are attached, without regard 
to their algebraic sign. 

19, If the initial line of an angle coincides with. OX, and its terminal 
line lies between OX and OY, the angle is said to be in the first quadrant; 
if the terminal line lies between OY and OX', the angle is said to be in 
the second quadrant; if between OX' and OY', in the third quadrant; if 
between Y' and OX, in the fourth quadrant. 

Thus, any positive angle between 0° and 90°, or 360° and 450°, or any 
negative angle between — 270° and — 360°, is in the first quadrant ; any 
positive angle between 90° and 180°, or 450° and 540°, or any negative 
angle, between — 180° and — 270°, is in the second quadrant. 

20. It follows from the definitions of § 17 that, for any angle in the 
first quadrant, all the functions are positive. 

It is also evident by inspection of the results of § 18 that : 

In the second quadrant f the sine and cosecant are positive, and the cosine, 
tangent, cotangent, and secant are negative. 

In the third quadrant, the tangent and cotangent are positive, and the sine, 
cosine, secant, and cosecant are negative. 

In the fourth quadrant, the cosine and secant are positive, and the sine, 
tangent, cotangent, and cosecant are negative. 



10 PLANE TRIGONOMETRY. 

It is usual to express the above iu tabular form, as follows : 



Functions. 


First 
Quad. 


Second 
Quad. 


Third 
Quad. 


Fourth 
Quad. 


Sine and cosecant 

Cosine and secant 

Tangent and cotangent .... 


+ 
+ 
+ 


+ 


+ 


+ 



21. Since the terminal lines of any two angles which differ by a mul- 
tiple of 360° are coincident (§ 13), it is evident that the trigonometric 
functions of two such angles are identical. 

Thus, the functions of 50°, 410°, 770°, - 310°, etc., are identical. 

22. Functions of 0° and 360°. 

Y 



a P(o,0) 



The terminal line of 0° coincides with the initial line OX. 
Let P be a point on OX such that OP = a. 
Then by § 16, the co-ordinates of P are (a, 0). 
Whence by definition, 



sin 0° = 5 = 0. 

a 



cos0 c 



1. 



tan C 
cotO 



0. 



sec0 c 
esc 0° 



By § 21, the functions of 360° are the same as those of C 
23. Functions of 90°. 

Y 

P (0,o) 

^90° 

— ^ X 



Let P be a point on OY such that OP = a. 
Then the co-ordinates of P are (0, a). 



TRIGONOMETRIC FUNCTIONS OF ANGLES IN GENERAL. H 

Whence by definition, 



sin 90° = - = 1. 

a 



tan 90° =^ = oo. 



sec90° = ^ = oo. 



cos90° = - = 0. 
a 



cot 90° = - = 0. 
a 



csc90° = - = l. 
a 



24. Functions of 180 c 



-P(-^O) 



180° 



o 



Let P be a point on OX' such that OP = a. 
Then the co-ordinates of P are (—a, 0). 
Whence by definition, 







sinl80°=- = 
a 



tan 180° = 

!sh -1. cot 180° = 

a 

25. Functions of 270°. 







= 0. 



= 00. 



270' 




O 
a 

P(0,-<0 



sec 180° = 



cscl80° = ^=oo. 



1, 



Let P be a point on OF such that OP = a. 
Then the co-ordinates of P are (0, — a). 
Whence by definition, 



sin270 c 



— a 



= -1. 



cos 270° = - = 0. 
a 



tan270° = — p = oo. 



cot 270° = — - = 0. 
— a 



sec 270° = ^=<x>. 




csc270 c 



— a 



= -1. 



Note. No absolute meaning can be attached to such a result as cdt0 c = ao ; it 
merely signifies that as an angle approaches 0°, its cotangent increases without limit. 
A similar interpretation must be given to the equations csc0°= oo, tan 90°= oo, etc. 



12 



PLANE TRIGONOMETRY. 



26. The results of the last four articles may be conveniently expressed 
in tabular form as follows : 



Angle. 
0° 


Sin; 


Cos. 


Tan. 


Cot. 


Sec. 


Cse. 





1 





oo 


1 


-■ 00 


90° 


1 





GO 





00 


1 


180° 





-1 





oo 


-1 


oo 


270° 


-1 





oo 





00 


-1 


360° 





1 





00 


1 


oo 



27. Functions of 120°, 135°, 150°, etc. 




Let OPM be a right triangle having OP, OM, and PM equal to 2, 1, 
and V3, respectively, and Z POM =60°. (Compare § 8.) 

Then Z XOP= 120°, and the co-ordinates of P are (- 1, V3). 
Whence by definition, 

V3 



sinl20 c 



cos 120° = 



tanl20° = -V3. 
cot 120° = 



1 — iva 



V3 



sec 120° = -2. 

cscl20° = -4: = ?V3. 
V3 3 



In like manner may be proved the remaining values given in the fol- 
lowing table, which are left as exercises for the student : 



Angle. 


Sin. 


Cos. 


rare. 


Co(. 


Sec. 


Csc. 


120° 


*V3 


_ 1 

2" 


-V3 


-*V3 


-2 


|V3 


135° 


1V2 


-iV2 


-1 


-l 


-V2 


V2 


150° 


1 

2 


-|V3 


-iV3 


-V3 


-|V3 


2 


210° 


1 

2 


-*V3 


iV3 


V3 


-|V3 


— 2 


225° 


- JVS 


-iV2 


1 


1 


-V2 


-V2 


240° 


-|V3 


-■_ i 

2" 


V3 


*vs 


_2 


-|V3 


300° 


-iv3 


1 

2 


-V3 


-1V3 


2 


-|V3 


315° 


-iV2 


iV2 


-1 


-1 


V2 


-V2 


330° 


_ i 

"2 


|V3 


-|V3 


-V3 


}V3 


-2 



TRIGONOMETRIC FUNCTIONS OF ANGLES IN GENERAL. 13 



28. Given the value of one function of an angle, to find the values of the 
remaining functions. (Compare § 6.) 

3 

1. Given sin A = ; find the values of the remaining functions of A. 

5 
The example may be solved by a method similar to that of § 6; since 
the sine is the ratio of the ordinate to the distance, we may regard the 
point of reference as having its ordinate equal to — 3, and its distance 
equal to 5. 

There are two points, P and P', which are 3 units below the axis of 
X, and distant 5 units from 0. 

Y 




PK-3) 



PU-3) 



There are then two angles, XOP and XOP', in the third and fourth 
quadrants, respectively, either of which may be the angle A. 

Now, OM = OM' = V&P - PM 2 = V25 - 9 = 4. 

Then the co-ordinates of P are (—4, — 3) : and of P', (4, — 3). 

Whence by definition : 



Angle. 


Cos. 


Tan. 


Cot. 


Sec. 


Csc. 


XOP 


_4 

5 


o 
O 

4 


4 
3 


5 

4 


5 
3 


XOP' 


4 
5 


3 
4 


4 
3 


5 
4 


5 
3 



Thus the two solutions to the problem are: 

cos A = T -, tan A = ± -, cot A = ± -, sec A = T -, csc A = — - : 
5 4 3 4 3 

where the upper signs refer to XOP, and the lower signs to XOP' 

2. Given cot A = 3 ; find the values of the remaining functions of A. 
The equation may be written either cot A = -, or cot A = — -• 



14 



PLANE TRIGONOMETRY. 



We may then regard the point of reference as having its abscissa 
equal to 3 and its ordinate equal to 1, or as having its abscissa equal to 
— 3 and its ordinate equal to — 1. 

There are two angles, XOP and XOP', in the first and third quad- 
rants, respectively, either of which satisfies the given condition. 




P'f-3,-1) 



Then OP= OP' = ^ OM 2 + PM 2 = V9~+1 =VM. 

Whence by definition : 



Angle. 


Sin. 


Cos. 


Tan. 


Sec. 


Csc. 


XOP 


1 

Vio 


3 
Vio 


1 
3 


Vio 
3 


Vio 


XOP 1 


l 
Vio 


3 
Vio 


1 

3 


Vio 
3 


-Vio 



Thus the two solutions are : 

sin^4=± , cos^4=± , tan A—-, sec.A=± , csc^l=±ViO. 

VIO VIO 3 3 

Note. It must be clearly borne in mind, in examples like the above, that the 
"distance" is always positive. 





EXAMPLES. 




In each of the following, find the values of the 


remaining functior 


3. sec A = -- 
4 


7. csc A = 

7 


11. tan^L = — 7 


4. cot^L = -— • 

5 


8. tan,4 = — • 
40 


12. csc A = 3. 


5. sin A = — 
17 


9. secA = --> 
2 


13. cos A = — 
b 


6. cosA = -?±. 
29 


10. sin A = --. 
5 


14. cot A = x. 



TRIGONOMETRIC FUNCTIONS OF ANGLES IN GENERAL. 15 



29. Functions of (—A) in terms of those of A 
Y 

P 




Fig. 1. 





1 


z 


PL 






N - 


A 


r 


fK 


y 


M 


\y 


°y ' 


P 


k ^ 


■rt 




Fig. 3. 

There may be four cases : A in the first quadrant (Fig. 1), J. in the 
second quadrant (Fig. 2), A in the third quadrant (Fig. 3), or A in the 
fourth quadrant (Fig. 4). 

In each figure, let the positive angle XOP represent the angle A, and 
the negative angle XOP' the angle — A. 

Draw PM perpendicular to XX', and produce it to meet OP' at P'. 

In the right triangles OPM and OP'M y the side OM is common, and 
APOM=£P'OM. 

Hence, the triangles are equal, and PM — P'M and OP 

Then in each figure, 



OP. 





abscissa P' = 


abscissa P, 






ordinate P' = - 


- ordinate P, 




d 


distance P' = 


distance P. 




Then, 








ord. P' 


ord. P ord. 7 J ' 


ord. P dist. P' _ 
abs. P abs. P' 


dist. P 


dist. P' 


dist. P abs. P' 


abs. P 


abs. P' 


abs. 7 J abs. P' 


abs. P dist. P' 


dist. P 


dist. P' 


dist. P ord. P' 


ord. P ord. P' 


ord.P 


Whence, 








sin (— ^4) = - 


-sin A tan(— A)—- 


-tan J.. sec (—^4) = 


sec A. ) 


cos (— A) — 


cosA cot (— A)=- 


- cot A. csc(—A)=- 


- esc A ') 



(1) 



16 PLANE TRIGONOMETRY. 

30. Functions of (90° + A) in terms of those of A 
P' Y P 





There may be four cases : A in the first quadrant (Fig. 1), A in the 
second quadrant (Fig. 2), A in the third quadrant (Fig. 3), or A in the 
fourth quadrant (Fig. 4). 

In each figure, let the positive angle XOP represent the angle A, and 
the positive angle XOP' the angle 90° + A. 

Take OP' = OP, and draw PM and P'M' perpendicular to XX'. 

Since OP is perpendicular to OP', and OM to PM', Z POM=Z OP'M'. 

Then the right triangles OPM and OP'M' have the hypotenuse and an 
acute angle of one equal to the hypotenuse and an acute angle of the other. 

Hence, the triangles are equal, and PM= OM' and 0M= P'M'. 

Then in each figure, 

ordinate P = abscissa P, 

abscissa P' = — ordinate P, 

and distance P' = distance P. 



Then, 



ord. P' abs. P 


abs. P' 


ord. P 


dist. P' dist. P 


ord. P' 


abs. P 


abs. P' ord. P 


dist. P' 


dist. P 


dist. P' dist. P 


abs. P' 


ord. P 


ord. P' abs. P 


dist. P'_ 


dist. P 



abs. P' 



ord. P 



ord. P' 



abs. P 



TRIGONOMETRIC FUNCTIONS OF ANGLES IN GENERAL. 17 

Or, sin (90° + A). = cos A cot (90° + A) = - tan A. -\ 

cos (90° -f- A) = - sin A sec (90° + A) = - esc A 1 (2) 

tan (90° + A) = - cot A. esc (90° + A) = sec A J 

31. The results of § 30 may be stated as follows : 

The sine, cosine, tangent, cotangent, secant, and cosecant of any angle are 
equal, respectively, to the cosine, minus the sine, minus the cotangent, minus 
the tangent, minus the cosecant, and the secant, of an angle 90° less. 

32. Functions of (90° — A) in terms of those of A. 

By § 31, sin (90° - A) = cos (- A) = cos A (§ 29). 

cos (90° - A) = - sin (- A) = sin A. 

tan (90° - A) = - cot (- A) = cot A. 

cot (90° -A)=- tan (- A) = tan A. 

sec (90° - A) = - esc (- A) = esc A. 

esc (90° - A)= sec (- A) = sec A 
These formulae were proved for acute angles in § 5. 

33. Functions of (180° — A) in terms of those of A. 

By § 31, sin (180° - A) = cos (90° - A) = sin ^. (§ 32). 
cos (180° - A) = - sin (90° - A) = - cos A 
tan (180° - A) = - cot (90° - A) = - tan A 
cot (180° - A) = - tan (90° -A)=- cot A 
sec (180° -A)=- esc (90° - A) = - sec A 
esc (180° - A) = sec (90° - ^4) = esc A 

34. By successive applications of the theorem of § 31, any function of 
a multiple of 90°, plus or minus A, may be expressed as a function of A. 

1. Express sin (270° 4- A) as a function of A. 

By § 31, sin (270° + A) = cos (180° + A)=— sin (90° + A) = - cos A 
If the multiple of 90° is greater than 270°, we may subtract 360°, or 
any multiple of 360°, from the angle, in accordance with § 21. 

2. Express sec (990° — A) as a function of A. 
Subtracting twice 360°, or 720°, from the angle, we have 

sec (990° -A) = sec (270° - A). 
And by § 31, sec (270° - A) = - esc (180° - A) = - esc A (§ 33). 

If the multiple of 90° is negative, we may add 360°, or any multiple 
of 360°, to the angle. 



18 PLANE TRIGONOMETRY. 

3. Express tan (— 180° + A) as a function of A. 
Adding 360° to the angle, we have 

tan (- 180° + A) = tan (180° + A). 
And by § 31, tan (180° + A) = - cot (90° + A) = tan A. 

EXAMPLES. 

Express each of the following as a function of A : 

4. sin(180° + J[). 9. sec (630° + A). 14. tan(- 450° - A). 

5. cos (270° -.4). 10. tan (-270° -A). 15. cos (- 900° - A). 

6. cot (450° + .4). 11. esc (-90°-^). 16. sin (810° - 4). 

7. esc (360° -^L). 12. cot (-180° + A). 17. esc (1080° + A). 

8. tan(540°-4). 13. sin (- 630° + A). 18. sec (1260° + A). 

35. By means of the theorem of § 31, any function of any angle, posi- 
tive or negative, may be expressed as a function of a certain acute angle. 

1. Express sin 317° as a function of an acute angle. 
By § 31, sin 317° = cos 227° = - sin 137° = - cos 47°. 

Since the complement of 47° is 43°, another form of the result is 
-sin43°(§ 5). 

Note. As in the examples of § 34, 360°, or any multiple of 360°, may be added 
to, or subtracted from, the angle. 

EXAMPLES. 

Express each of the following as a function of an acute angle : 

2. cos 322°. 4. sec 559°. 6. cot (- 378°). 

3. tan208°. 5. esc 803° 45'. 7. sin (- 139° 5'). 

It is evident from the above that any function of any angle can be 
expressed as a function of a certain acute angle less than 45°. 

Express each of the following as a function of an acute angle less than 

45°: 

8. cot 155°. 10. sec 457°. 12. tan (-681°). 

9. sin 1138° 36'. 11. cos 496° 20'. 13. esc (-257°). 

14. Find the value of esc (- 210°). 
Adding 360° to the angle, we have 

esc (-210°)= esc 150°. 
And by § 31, esc 150° = sec 60° = 2 (§ 8). 

Find the values of the following : 

15. cot 405°. 17. esc 600°. 19. cos (-420°). 

16. sin 480°. 18. tan 690°. 20. sec (-225°). 



GENERAL FORMULA. 



19 



III. GENERAL FORMULAE. 

36. It follows immediately from the definitions of § 17 that, if x is 
any angle, 

1 ... 1 1 



sin a* 



cos a; 



esc x 
1 



sec a? 
37. To prove the formula 





sec x = 

cos a 


cot a; 


1 

cot X = 

tana- 


esc x = — — 

sina; 


. n „„ sina- 

tana = 





cos a 



I. When the angle x is acute (or in the first quadrant). 




Fig. 1. 



In the right triangle ABC, let BAC be the angle x. 

BC 
BC AB sin x 



By § 2, 



tan x = 



AC AC cos x 
AB 

II. When x is in the second, third, or fourth quadrant. 



0) 



(4) 




y, yf 

Fig. 2. Fig. 3. 

In each figure, let the positive angle XOP represent the angle x, am 
d raw PM perpendicular to XX'. 

Then in each figure, by the definitions of § 17, 

ord. P 
ord. P dist. P sin x 



tana? 



abs. P abs. P cos a? 
dist. P 



20 PLANE TRIGONOMETRY. 

38. To prove the formula 





, cos a; 
cot X = — 

sma; 


By (3), § 36, cota; = 


1 1 

" tan x sin x 




cos a; 


39. To prove the formula 





(§37) = «5E. 
J sin x 



tv a- v, no 2 PM~ , OM 1 

Dividing by OP , ^=^ + ^^ = 1. 

OP 2 OP 2 

PM 2 OM 1 

But in either figure, 7 = sin 2 a?, and 



(5) 



sin 2 x -f- cos 2 x — 1. (6) 

Note. Sin 2 as signifies (sin a;) 2 ; that is, the square of the sine of x. 

I. When the angle # is acute (or in the first quadrant). 
In Fig. 1, § 37, we have by Geometry, 

BC 2 + AC 2 = AB*. 

»**.,*. (gy + (^)'=.. 

Then by definition, (sin as) 2 + (cos a?) 2 = 1. 
That is, sin 2 x + cos 2 £ = 1. 

II. When x is in the second, third, or fourth quadrant. 
In each of the figures 2, 3, and 4 of § 37, we have 

PM 2 + OM 2 = OP 2 . 



0P - 0p ; 

Whence, sin 2 x -f- cos 2 x = 1. 

Formula (6) may be written in the forms 

sin 2 a; = 1 — cos 2 jc, and cos 2 x = 1 — sin 2 a?. 

40. To prove the formulae 

sec 2 cc = 1 + tan 2 x, (7) 

aw d esc 2 a) = 1 + cot 2 x. ( 8 ) 

By (6), 1 = cos 2 a; + sin 2 #. (A) 

Dividing by cos 2 x, — — = 1 H — 

cos- a; cos 2 a; 

Whence by (3) and (4), sec 2 x = 1 + tan 2 x. 



GENERAL FORMULA. 



21 



Again, dividing (A) by sin 2 x, we have 

1 _ -j cos 2 a; 
sin 2 x sin 2 x 

Whence by (3) and (5), csc 2 # = 1 + cot 2 a;. 

41. To find the values of sin (x -f- y) and cos (x -f- y) in terms of the sines 
and cosines of x and y. 

I. When x and y are acute, and x + y acute. 

c/ 




Let A OB and BOC denote the angles x and y, respectively. 

Then, Z AOO= x + y. 

"From any point (7 in 0(7 draw CA and (75 perpendicular to OA and 
02* ; and draw BD and 5^ perpendicular to OA aiuj J. (7. 

Since EC and 5(7 are perpendicular to OA and 05, the angles BCE 
and ^105 are equal ; that is, Z BCE = x. 



and 



Now, 
But, 

i 

Whence, 
Again, 

But, 



and 



■ , . , .4(7 BD+CE BD . CE 

sm (a? 4- ?/) — = ! = 

V J) OC OC OC OC 

BD BD OB 

= — — x = sm x cos y, 

OC OB OC 3i 

CE CE BC 

= x = cos x sin y. 

OC BC OC y 

sin (x + y) = sin x cos y + cos x sin y. 

OA OD-BE OD BE 



cos (x -f- y) 



OC 



OC 



OC OC 



OD OD OB 

= x = cos x cos y, 

OC OB OC *' 

BE BE BC ... 

0C = BC X 0C = SinXSmy ' 



(9) 



Whence, cos (x + y) = cos x cos y — sin x sin y. 



(10) 



PLANE TRIGONOMETRY. 

II. When x and y are acute, and x + y obtuse. 




.Let DOB and BOC denote the angles x and y, respectively. 

Then, ZDOC=x + y. 

From any point C in OC draw CB perpendicular to OB, and CA 
perpendicular to DO produced; and draw BD and BE perpendicular 
to OD and AC. 

Since EC and BC are perpendicular to OD and OB, the angles BCE 
and DOS are equal ; that is, Z BCE = x. 



Then by §17, sin DOC 



AC BD + CE BD , CE 



OC 



OC 



C + 



But, 



and 



Whence, 
Again, 



BD BD OB 

= x = sin x cos y, 

OC OB OC y ' 

CE CE BC 

- — - = x = cos x sin y. 

OC BC OC y 

sin (x -\-y) = sin x cos y + cos x sin y. 

OA = OD-BE = OD BE 

~ OC OC 



cos DOC 



But, 



and 



OC OC 

0» = 0B 0B = 

OC OB OC U) 

BE BE BC 

-—— = x — — = sin x sin y. 

OC BC OC U 



Whence, cos (x + y) = cos x cos y — sin x sin y. 

42. Formulae (9) and (10) are very important, and it is necessary to 
prove them for all values of x and y. 

They have already been proved when x and y are any two acute 
angles ; or, what is the same thing, when they are any two angles in the 
first quadrant. 

Now let a and b be any assigned values of x and y, for which (9) and 
(10) are true. 



GENERAL FORMULA. 



23 



By (2), § 30, sin [90° + (a + &)] = cos (a + b), 
and cos [90° + (« + &)]=_ sin (a + 6). 

Whence, by (9) and (10), 

sin [90° + (a + &)] = cos a cos 6 — sin a sin 6, (A) 

and cos [90° + (a + &)] = — sin a cos 6 — cos a sin 6. (B) 

But by (2), § 30, cos a = sin (90° + a), and — sin a — cos (90° + a). 
Then, (A) and (B) may be written in the forms 

sin [(90° + a) + &] = sin (90° + a) cos & + cos (90° + a) sin b, 
and cos [(90° + «) + &] = cos (90° + a) cos b — sin (90° + a) sin 6 ; 

which are in accordance with (9) and (10). 

Therefore, if (9) and (10) hold for any assigned values of x and y, 
they also hold when one of the angles is increased by 90°. 

But they have been proved to hold when x and y are both in the 
first quadrant; hence, they hold when x is in the second quadrant and 
y in the first. 

And since they hold when x is in the second quadrant and y in the 
first, they hold when x and y are both in the second quadrant ; and so on. 

Hence, (9) and (10) hold for any values of x and y whatever, positive 
or negative. 



43. By (9), sin [x 4- (— ?/)] = sin x cos (— y) + cos x sin (— y) 

= sin x cos y -f cos x(— sin y), by (l), § 29. 
Whence, sin (x — y) = sin x cos y — cos x sin y. (11) 

By (10)> cos [x + (— y)~\ = cos x cos (— y) — sin x sin (— y) 

= cos x cos y — sin x (— sin ?/). 
cos (x — y) = cos x cos ?/ 4- sin x sin ^/. 

sin (x + y) 



Whence, 
44. By (4), 



(12) 



tan (a; -f- y) 



cos (a; + y) 



sin x cos w + cos x sin v , /rtX -, ,_ rtX 

= — ■ — : : — -, by (9) and (10). 

cos x cos y — sin x sin y 



Dividing each term of the fraction by cos x cos ?/, 



tan (x +- y) = 



sm # cos ?/ cos a; sin y 
cos # cos y cos a; cos # 



cos x cos # _ sin x sin y 
cos x cos 2/ cos # cos y 
tan # + tan y 
1 — tan x tan y 



(13) 



24 PLANE TRIGONOMETRY. 

In like manner, we may prove 

+„„ /„. \ tan x — tan y ,_., 

tan (a; — ?/) = £-. (14 ) 

1 + tan x tan y 

Again, by (5), cot (* + y) = C ° S ^ + ^ 

6 ' J w ' v -r ^; sin (a? -f y) 

_ cos a? cos ?/ — sin x sin y 
"sin a; cos y + cos a; sin y 

Dividing each term of the fraction by sin x sin y, 



cot (a + ?/) 



cos x cos y _ sin x sm ?/ 
sin # sin y sin a? sin y 



sin a? cos 



2/ cos # sin ?/ 



(15) 



sin x sm 2/ sin x sin 2/ 
_ cot x cot ?/ — 1 
cot y + cot x 

In like manner, we may prove 

, , s cot a? cot y + 1 /n _ 

cot (a; — ?/) = ^— ! (16 > 

cot y — cot a; 

45. From (9), (10), (11), and (12), we have 

sin (a + ft) = sin a cos 6 + cos a sin ft. (A) 

sin (a — ft) = sin a cos 6 — cos a sin ft. (B) 

cos (a + 5) = cos a cos ft — sin a sin ft. (C ) 

cos (a — ft) = cos a cos ft + sin a sin ft. (D) 

Adding and subtracting (A) and (B), and then (C) and (D), 

sin (a -f- ft) + sin (a — ft) = 2 sin a cos ft. 

sin (a -f- ft) — sin (a — ft) = 2 cos a sin ft. 

cos (a + ft) + cos (a — ft) = 2 cos a cos ft. 

cos (a H- ft) — cos (a — ft) = — 2 sin a sin ft. 

Let a + ft = #, and a — ft = y. 

Then, a — %(x + 2/)? an d & = -| (a; — y). 

Substituting these values, we have 

sin x 4- sin y = 2 sin i (a; + y) cos i (a? — y). (17) 

sin # — sin y = 2 cos £ (a; + y) sin i (a? — y). (18) 

cos # -f- cos y — 2 cos \(x + y) cos ^ (# — y). (19 ) 

cos a; — cos y — — 2 sin J (ar + y) sin i (a?. — ?y). (20) 



GENERAL FORMULA. 



25 



46. By (17) and (18), we have 

sin x 4- sin y _ 2 sin \ (x + y) cos %(x — y) 
sin x — sin ?/ — 2 cos -|- (a? + y) sin i (a? — y) 

= tan i (x 4- ?/) cot -J- (x — y) 

tan i(aj + ?/) 
- tanl(,- y y b ^ 3 > 



(21; 



47. By (9) and (11), we have 

sin (x + y) sin (x — y) = (sin a; cos # + cos x sin ?/) (sin a? cos y — cos # sin y) 
= sin 2 x cos 2 y — cos 2 x sin 2 y 
= sin 2 a* (1 — sin 2 y) — (1 — sin 2 a?) sin 2 y (§ 39) 
= sin 2 x — sin 2 x sin 2 y — sin 2 ?/ + sin 2 x sin 2 ?/ 
= sin 2 x — sin 2 ?/. 
The result may also be written 

sin (x + ?/) sin (x — y) = 1 — cos 2 a? — (1 — cos 2 y) (§ 39) 
= cos 2 y — cos 2 x. 
In like manner, we may prove 

cos (x 4- y) cos (x — y)= cos 2 a; — sin 2 # 

48. Functions of 2 a. 

Putting y = x in (9), we have 



cos 2 y — sin 2 x. 



sin 2 a? = sin x cos a; 4- cos x sin a; 
== 2 sin x cos a\ 
Putting y = x in (10), we obtain 

cos 2 a; = cos 2 x — sin 2 a*. 
We also have by § 39, 

cos 2 x =(1 — sin 2 a;)— sin 2 # = 1—2 sin 2 a;, 
and cos 2 x = cos 2 a; — (1 — cos 2 a;)= 2 cos 2 a; — 1. 

Putting y = x in (13) and (15), we have 

2 tan a? 



tan 2 a; = 
cot 2x = 



1 — tan 2 a?' 
cot 2 x — 1 



2 cot a; 

49. Functions of \x. 

From (27) and (28) we have, by transposition, 

2 sin 2 a; = 1 — cos 2 x, and 2 cos 2 a; = 1 4- cos 2 x. 



(22) 

(23) 
(24) 



(25) 

(26) 

(27) 
(28) 

(29) 
(30) 



26 .PLANE TRIGONOMETRY. 

Putting ^ x in place of x, and therefore x in place of 2 x, we have 

2 sin 2 i# = 1 — cos x, (31) 

2 cos 2 -J- a = 1 -f- cos x. (32) 

Again, putting \x in place of a; in (25), 

2 sin,V#cos^#= sin a;. (A) 

Dividing (31) by (A), we have, by (4), 

. , 1 — cos a? ,__v 

tan 4 a; = — (33) 

sin a; 

Dividing (32) by (A), cot ±x = 1+ . cosa? . (34) 

50. Functions of 3x. 

We have, sin 3 x = sin (2 a; + a;) = sin 2 a; cos x + cos 2 a; sin a?, by (9) 
= (2 sin a; cos a;) cos a; +(1—2 sin 2 a;) sin a; (§ 48) 
= 2 sin x (1 — sin 2 a?)+ sin a; — 2 sin 3 a; (§ 39) 
= 3 sin x — 4 sin 3 a?. (35) 

Also, cos 3 x = cos (2 # + x) = cos 2 x cos a; — sin 2 a? sin jc, by (10) 

= (2cos 2 a;— l)cosa;— (2 sin a? cos a?) sin a? (§ 48) 
= 2 cos 3 a; — cos a? — 2 cosa;(l — cos 2 a;) (§ 39) 
= 4 cos 3 x — 3 cos x. (36) 

Again, tan 3 x = tan (2 a; + x) = — — — ^-J — ^-, by (13) 

1 — tan 2 a? tan a; 

2 tan a? , , 

tana; 



1— tan 2 a; 

-, by (29) 



1- 



/ 2 tana; \, ' 

[ — ] tan x 

\1 — tan 2 xj 



_ 2 tan x-\-(l — tan 2 x) tan x _ 3 tan a; — tan 3 a; 
1 — tan 2 x — 2 tan 2 a; 1 — 3 tan 2 a; 



(.37) 



EXERCISES. 

51. 1. Prove the relation sec 2 a; esc 2 a; = sec 2 a; -f- esc 2 a;. 

^ 2 

cos 2 a; 



By (3), sec 2 * cse'x = , 1 , , = <**'* + <j°f* t by (6) 



cos" x sur x cos z x sin- a; 

+ -— — = sec 2 a; + esc 2 x. 



cos" a; sin" x 



GENERAL FORMULAE. 27 

ft t» i.v -i 4.- sin 3 x — sin x , 

2. Prove the relation = tan x. 

cos 3 a; + cos a; 

~ , \ i / x sin 3 x — sin x 2 cos 1(3 x + #) sin 4- (3 x — x) 

By (18) and (19), 5 — ■ =^ , ,0 \ i Vk> (=tanx. 

. v ' cos 3 a; -f- cos # 2 cos £ (3 a? -f a?) cos £ (3 a; — a;) 

3. Prove the relation tan (a; + y) - tan a; =tany 

1 4- tan (x + 1/) tan a? 

-u r-.v tan (a; + y) — tan a; . r/ . N -, , 

By (14), * — X -^ LL — = tan \(x + y) — x\ = tan y. 

J v /' 1 + tan (x + y) tan aj LV ^ J * 

Prove the following relations : 

4 sin (x + y) _ tan x 4- tan y - cos (a; 4- y) _ 1 — tan x tan y 

sin (a? — y) tan a; — tan y cos (x — y) 1 4- tan a? tan y 

fi cos x 4- cos y n . 1 / , n _, 1 / s 

6. -!- ^ = -cot-i-(a:-hy)coti-(a;-y). 

cos a; — cos y 

7. sin (a; + y + z) = sin x cos y cos z + cos a; sin y cos z 

4- cos x cos y sin z — sin x sin y sin z. 

8. cos (a; 4- V + z) = c °s % cos y cos z — sin a; sin y cos z 

— sin x cos y sin z — cos x sin y sin z. 

9. tan(60 o +a}--cot(30°-a;)=0. 12 / tan a; 4- 1Y = 1 + sin 2 a; 

\tan x — l) 1 — sin 2 # 

10 — ¥~~a o = sec 2 A sin 5 a; 4- sin a; , n Q 

esc 2 J. — 2 13. = tan 3 x. 

cos o x 4- cos a; 

- , sin 2 a? cos 2x + A sin 3 a; — sin 5 a; , . 

]1. — : =seca\ 14. = — cot 4 a:. 

sin x cos x cos ox- cos 5 a; 

15. sin 4 x = 4 sin a? cos x — 8 sin 3 a? cos a;. 

16. cos 4 a; ='1 — 8 cos 2 x 4- 8 cos 4 x. 

17. By putting a; — 45° and y = 30° in (11) and (12), prove 

sin 15° = \ ( V6 - V2), cos 15° = \ ( V6 + V2). 

18. By putting x = 30° in (33) and (34), prove 

tan 15° = 2 - V3, cot 15° = 2 4- VS. 

19. Using the results of Ex. 17, prove 

sec 15° = V6 - V2, esc 15° = V6 4- V2. 

20. By putting x = 45° in (31) and (32), prove 

sin 221° = ^V2-V2, cos 22^° = iV2+V2. t 



28 PLANE TRIGONOMETRY. 

21. By putting x = 45° in (33) and (34), prove 

tan 221° = V2 - 1, cot 22i = V2 + 1. 

22. By putting x = 221° in (7) and (8), and using the result of 
Ex. 21, prove 



sec 



221° = V4 - 2 V2, esc 22i° = V4 + 2 V2. 



'2 

Prove the following relations : 

23. tan (45° + x) - tan (45° - x) = 2 tan 2 x. 

24. cos 4 a; — sin 4 x = cos 2 x. 25. ■ = cot \ x. 

esc a? — cot x 

26. sin 2 (x-\-y) — sin 2 (x — y) = sin 2 a; sin 2 1/. 

27. tm(»-y) + tanv = fama . 
1 — tan (a? — y) tan ?/ 

28. cos 5 ^4 cos 3 A + sin 5 A sin 3 ^L = cos 2 A 

29. sin {A + 5) cos (^ - B) - cos (^1 + B) sin ( JL - B) = sin 2 B. 

OA cos 3 a; . sin 3 x , 00 sin 4 a; + sin 3 a; , , 

30. — 1 = 2 cot 2 a;. 33. — = cot 1 x. 

sm a; cos x cos 3 a; — cos 4 a; 

31. sin 2a; = 2 tan * . 34. sin 50° + sin 10° = sin 70°. 

1 + tan 2 a; 

00 o 1 — tan 2 a? OK sin a; + sin 2x . 

32. cos 2 a; = — 35. — — - = tan x. 

1 + tan 2 x 1 + cos x + cos 2 a; 

36. 2 cos 3 a? sin x = sin 4 x — sin 2 a;. 

37. cos 5 x = 5 cos a? — 20 cos 3 x 4- 16 cos 5 a;. 
cos a; cot 1 a; -}- 1 Q q 4- >• 4 tan x — 4 tan 3 a; 



38. „ ^°. = ""7-*-r^. 39. tan 4 a; 

2 



1 — sin x cot i a; — 1 1 — 6 tan 2 x + tan 4 a; 



40. (sin x + cos a?) (2 — sin 2 a?) = 2 (sin 3 a; + cos 3 x). 

41. (sin a; — sin 2/) 2 + (cos x — cos yf = 4 sin 2 ~ ^ - 

42 l + sm iL -cos 5=tan 43 sin3x-cos.3x =2sin2a ,_ 1 

1 + sin a; + cos a; sin a; + cos x 



MISCELLANEOUS THEOREMS. 29 



IV. MISCELLANEOUS THEOREMS. 

52. Circular Measure of an Angle. 

An angle is measured by finding its ratio to another angle, adopted 
arbitrarily as the unit of measure. 

The usual unit of measure for angles is the degree, which is an angle 
equal to the ninetieth part of a right angle. 

Another method of measuring angles, and one of great importance, is 
known as the Circular Method; in which the unit of measure is the angle 
at the centre of a circle subtended by an arc whose length is equal to the 
radius. 




Thus, let A OB be any angle; and let AOC be the unit of circular 
measure; that is, the angle at the centre subtended by an arc whose 
length is equal to OA. 

ZAOB 



Then, circular measure A OB 

But by Geometry, 



ZAOC 
AOB zltcAB arcAB 



ZAOC arc .40 OA 

Whence, circular measure A OB = 

\ OA 

That is, the circular measure of an angle is the ratio of its subtending 
arc to the radius of the circle. 

53. By § 52, the circular measure of a right angle is the ratio of one- 
fourth the circumference to the radius. 

But if R denotes the radius, the circumference of the circle is 2 irR. 

Whence, circular measure of 90° = * of 2 7rR = -• 

R 2 

It follows from the above that the circular measure of 180° is tt ; of 
60°, | ; of 45°, j; etc. 

That is, an angle expressed in degrees may be reduced to circular meas- 
ure by finding its ratio to 180°, and multiplying the result by ir. 

Thus, since 115° is — of 180°, the circular measure of 115° is — - 
36 36 



30 PLANE TRIGONOMETRY. 

54. Conversely, an angle expressed in circular measure may be reduced 
to degrees by multiplying by 180° and dividing. by ir; or, more briefly, by 
substituting 180° for w. 

Tims, — = — of 180° == 84°. 
' 15 15 

55. In the circular method, such expressions may occur as " the angle 
|," "the angle 1," etc. 

These refer to the unit of circular measure ; thus, the angle J signifies 
an angle whose subtending arc is two-thirds of the radius. 

The angle 1, that is, the angle whose subtending arc is equal to the 
radius, or the unit of circular measure, reduced to degrees by the rule of 



54, gives 



57.2958°, approximately. 



7T 3.14159... 
Then the rule of § 54 may be modified as follows : 

An angle expressed in circular measure may be reduced to degrees by 
multiplying by 57.2958°. 

Thus, the angle f = f x 57.2958° = 38.1972° = 38° 11' 49.92". 

EXAMPLES. 
56. Express each of the following in circular measure : 



1. 120°. 


3. 


67° 30'. 


5. 


86° 24'. 


7. 


163° V 30". 


2. 315°. 


4. 


146° 15'. 


6. 


53° 20'. 


8. 


88° 53' 20". 


Express 


each of the following 


in degree measure : 






9 ' T" 




"• it 




»j- 




15. -"I 


10 11^ 
24 




12. | 




»i- 




16. 3?r + 2 
5 



57. Inverse Trigonometric Functions. 

The expression sin -1 x, called the inverse sine of x, or the anti-sine of 
x, signifies the angle whose sine is x. 

Thus, the statement that the sine of the angle x is equal to y may be 
expressed in either of the ways 

sin x = y, or x = sin -1 y. 
In like manner, cos -1 x signifies the angle whose cosine is x ; tan -1 x, 
the angle whose tangent is x ; etc. 

Note. The student must be careful not to confuse the above notation with the 
exponent — 1 ; the — 1 power of sinx is expressed (sinx) -1 , and not sin -1 a. 



MISCELLANEOUS THEOREMS. 



31 



It is evident that the sine of the angle whose sine is x is x ; that is, 
sin (sin -1 x) = x. 

In like manner, cos (cos -1 x) = x ; tan (tan -1 x) = x ; etc. 

58. By aid of the principles of § 57, we may derive from any formula 
involving direct functions a relation between inverse functions. 

1. From the formula tan (x + y) = ' — — ^-, prove 

1 — tan x tan ?/ 



Let 

Then by § 51 



tan- 1 a + tan" 1 b = tan" 1 a + h . 
1 — ab 

tana? •= a, and tany = fr. 
x = tan -1 a, and ?/ == tan -1 6. 



Substituting these values in the given formula, 



Whence, 



tan (tan- 1 a + tan" 1 b) = a + b . 
1 — ab 

tan" 1 a + tan" 1 b = tan" 1 a + b . 
1 — ab 



2. Prove the relation cot -1 a — sec -1 b — cos -1 a — 

b Va 2 4- 1 
Let cot -1 a = a, and sec -1 b = y. 

Then, cot x = a, and sec y = b. 

Now, cos (x — y) = cos x cos y -f- sin x sin y. 

To find the values of the sines and cosines of x and y, we may use the 
method of § 6. 



(A) 





In the right triangle containing the angle x, the adjacent side is a, 
and the opposite side 1 ; then, the hypotenuse is Va 2 -f 1. 

In the right triangle containing the angle y, the hypotenuse is b, and 
the adjacent side 1 ; then, the opposite side is V& 2 — 1. 

Substituting the values of cos x, cos y, sin x, and sin y in (A), we have 

cos (a -2/) 



Va 2 + 1 6 Va 2 + 1 



VP^l a+V6 2 



6 Va 2 + 1 



Whence, x — y or cot" 1 a — sec" 1 b = cos" 1 a + 

b Va 2 + 1 



32 PLANE TRIGONOMETRY. 



EXAMPLES. 

cot 2 X — 1 

3. From the formula cot 2 x = — , prove 

2 cot a r 

2 cot" 1 a = cot" 1 **-=lI. 

2a 

4. From the formula cos 2 x = 1 — 2 siu 2 a*, prove 

2 sin" 1 a = cos" 1 (1-2 a 2 ). 

5. From the formula sin 2 a* = 2 sin it* cos x, prove 

2 cos" 1 a = sin" 1 (2 aVl - a 2 ). 

6. From the formula cos (x -\- y) = cos a* cos y — sin a* sin y, prove 



cos -1 a + cos -1 b = cos -1 (a& — Vl — a 2 Vl — & 2 ). 

7. From the formula sin 3 x = 3 sin a* — 4 sin 3 x, prove 

3 sin -1 a = sin -1 (3 a — 4 a 3 ). 

Prove the following relations : 

8. cot" 1 a + cot- 1 6 - cot- 1 06 " 1 . 

a + b 

9. 2 cos -1 a = cos -1 (2 a 2 — 1). 



10. sin -1 a — sin -1 b = sin -1 (aVl — b 2 — 6 Vl — a 2 ). 

11. 3tan- 1 o = tan- 1 ? a ~ a / 

1 — 3a- 

_ 1 a 2 -6 2 + l 



12. cot -1 (a — b) — cot -1 (a + 6) = cot 



2& 



.„ . , u , ,ah + Vl — a 2 Vl — b 2 

13. sin" 1 a + cos -1 b = tan -1 — ^— 

&VT^o 2 -aVl-& 2 

i>i i -it. ^Va^^T+V^l 

14. sec -1 a — esc x b = cos x ! 

ab 



ie i il . _ia-hVa 2 — 1 

15. tan _] a + cos -1 - — sin x — - 

a a Va 2 + 1 

16. tan- 1 — °^- - tan" 1 ^+^ = tan" 1 -L. 

a — 1 a 2 a- 



iiv n • -l *. _i 2 a Vl — a 2 

17. 2 sin' 1 a = tan" 1 — — — — — - 

1 — 2a- 

18. tan- 1 a+2tan- 1 6 = tan~ 1 ^L^!l±l^ 

l — b 2 — 2ab 



MISCELLANEOUS THEOREMS. 



33 



59. The following table expresses the value of each of the six prin- 
cipal functions of an angle in terms of the other five : 



sin A 
cos A 
tan A 
cot A 

sec .4 
esc A 



sin A 





Vl-cos 2 vl 


Vl-sin 2 ^l 




sin A 

Vl — sin 2 A 


Vl — cos 2 J. 
cos .4 


Vl — sin 2 ^4 


cos A 


sin A 
1 


Vl— cos 2 ^4 

1 






Vl — sin 2 ^l 

1 


cos A 

1 







Vl - cos 2 A 



t&nA 



Vl + tan 2 A 

1 



Vl + tan 2 A 



tan A 



Vl+tan 2 J. 

VT+tan 2 ^! 



tan A 



cot A 



Vl + cot 2 ^4 



1 

Vl + cot 2 ^l 
cot J. 


Vsec 2 ^l — 1 
sec A 

1 


Vl + cot 2 ^4 
1 


sec A 


cot A 


Vsec 2 A— 1 
1 




Vsec 2 ^l-1 


Vl + cot 2 .4 





sec^L 



Vsec 2 ^!— 1 



1 
esc A 



Vcsc 2 yl-1 



cscvl 

1 



Vcsc 2 ^4— 1 



Vcsc 2 ^l— 1 
esc A 



Vcsc 2 ^4 — 1 



The reciprocal forms were proved in § 36. 

The others may be derived by aid of §§ 36, 37, 38, 39, and 40, and are 
left as exercises for the student. 

As an illustration, we will give a proof of the formula 



cos A — 



Vcsc 2 ^! - 
esc A 



By § 39, cos A = Vl - sin 2 A=\ll \- = a/cs(?2 ^ ~ 1 . 

* ese 2 A cse A 



csc^L 



They may also be conveniently proved by the method of § 6 ; thus, let 
it be required to prove the formula for each of the other functions in 
terms of the secant. 

sec .4 



We have 



sec A = 




Since the secant is the ratio of the hypotenuse to the adjacent side, we 
take AB = sec A, and AC = 1 ; whence, BC = V ' Al£— A(f— Vsec 2 ^. — 1. 



34 



PLANE TRIGONOMETRY. 



Then by definition, 



sin ^4 
cos A 



V sec- A — l 



sec^l 



sec A 



tan A = VsecM — 1, 
cot A= — , 



esc A = 



sec A 



VsecM — 1 



VsecM - 1 
60. Line Values of the Functions. 

Y 

F 





J 


T 
C 






o\ d\ 


A X 


1 i 


B/n 


E 



Y 

Fig. 4. 



Let AOB be any angle. With as a centre, and a radius equal to 1, 
describe the circle AB ; draw BD and AE perpendicular to XX', and CF 
perpendicular to YY'. Then by § 17, the functions of AOB are : 





Sin. 


Cos. 


Tan. 


Cot. 


Sec. 


Csc. 


Fig. 1. 
Fig. 2. 
Fig. 3. 
Fig. 4. 


BD 
OB 
BD 
OB 
BD 
OB 
BD 
OB 


OD 
OB 
OD 
OB 
OD 
OB 
OD 
OB 


BD 
OD 
BD 
OD 
BD 
OD 
BD 
OD 


OD 
BD 
OD 
BD 
OD 
BD 
OD 
BD 


OB 
OD 
OB 
OD 
OB 
OD 
OB 
OD 


OB 
BD 

OB 
BD 
OB 
BD 
OB 
BD 



MISCELLANEOUS THEOREMS. 



35 



But since the right triangles OBD, OEA, and OCF are similar, and 
OA = OC = 1, we have 



BD AE Aw 

OD = CF ==CF 
BD OC 



OB OE nv 
OD == OA =OE > 



°2=°l = OF 
BD OC 



Whence, since OB= 1, the functions of AOB are 





Sin, 


Cos. 


Tan. 


Cot. 


Sec. 


Csc. 


Fig. 1. 


BD 


OD 


AE 


CF 


OE 


OF 


Fig. 2. 


BD 


-OD 


-AE 


-CF 


-OE 


OF 


Fig. 3. 


-BD 


-OD 


AE 


CF 


-OE 


-OF 


Fig. 4. 


-BD 


OD 


-AE 


-CF 


OE 


-OF 



That is, if the radius of the circle is 1, 

The si?ie is the perpendicular drawn to XX' from the intersection of 
the circle with the terminal line. 

The cosine is the line drawn from the centre to the foot of the sine. 

The tangent is that portion of the geometrical tangent to the circle at 
its intersection with OX included between OX and the terminal line, pro- 
duced if necessary. 

The cotangent is that portion of the geometrical tangent to the circle 
at its intersection with OY included between OY and the terminal line, 
produced if necessary. 

The secant is that portion of the terminal line, or terminal line pro- 
duced, included between the centre and the tangent. 

The cosecant is that portion of the terminal line, or terminal line pro- 
duced, included between the centre and the cotangent. 

And with regard to algebraic signs, 

Sines and tangents measured above XX' are positive, and below, nega- 
tive; cosines and cotangents measured to the right of YY ! are positive, 
and to the left, negative ; secants and cosecants measured on the terminal 
line itself are positive, and on the terminal line produced, negative. 

The above are called the line values of the trigonometric functions. 

They simply represent the values of the functions when the radius is 
1 ; that is, the numerical value of the sine of an angle is the same as the 
number which expresses the length of the perpendicular drawn to XX' 
from the intersection of the circle and terminal line. 



36 



PLANE TRIGONOMETRY. 



61. To trace the changes in the six principal trigonometric functions of 
an angle as the angle increases from 0° to 360°. 




Let the terminal line start from the position OA, and revolve about 
the point as a pivot, in a direction contrary to the motion of the hands 
of a clock. 

Then since the sine of the angle commences with the value 0, and 
assumes in succession the values I?iA> B 2 D 2 , OO, B 3 D 3 , BJ)±, etc. (§ 60), 
it is evident that, as the angle increases from 0° to 90°, the sine increases 
from to 1 ; from 90° to 180°, it decreases from 1 to ; from 180° to 270°, 
it decreases (algebraically) from to —1; and from 270° to 360°, it in- 
creases from — 1 to 0. 

Since the cosine commences with the value OA, and assumes in suc- 
cession the values OA, OD 2 , 0, -OD 3 , -OD 4 , etc., from 0° to 90°, it. 
decreases from 1 to ; from 90° to 180°, it decreases from to — 1 ; from 
180° to 270°, it increases from -1 to 0; and from 270° to 360°, it increases 
from to 1. 

Since the tangent commences with the value 0, and assumes in suc- 
cession the values AE X , AE 2 , x, — AE 3 , — AE A , etc., from 0° to 90°, it 
increases from to oc ; from 90° to 180°, it increases from — oo to ; from 
180° to 270°, it increases from to oo ; and from 270° to 360°, it increases 
from — co to 0. 

Since the cotangent commences at oo, and assumes in succession the 
values CF X , CF 2 , 0, - OF 3 , - CF 4 , etc., from 0° to 90°, it decreases from 
oo to ; from 90° to 180°, it decreases from to - oo ; from 180° to 270°. 
it decreases from oo to 0; and from 270° to 360°, it decreases from to 



Since the secant commences with the value OA, and assumes in suc- 
cession the values OE x , OE 2 , oo, — OE 3 , — OE A , etc., from 0° to 90°, it 
increases from 1 to oo ; from 90° to 180°, it increases from — oo to — 1 ; 
from 180° to 270°, it decreases from - 1 to - oo ; and from 270° to 360°, 
it decreases from oo to 1. 



MISCELLANEOUS THEOREMS. 37 

Since the cosecant commences at oc, and assumes in succession the 
values OF,, OF,, 00, OF 3 , OF,, etc., from C to 90°, it decreases from w 
to 1 ; from 90° to 180°, it increases from 1 to oo ; from 180° to 270°, it 
increases from — oo to — 1 ; and from 270° to 360°, it decreases from — 1 
to — 00. 

Note. Wherever the symbol oo occurs in the above discussion, it must be inter- 
preted as explained in the Note to § 25. 

62. Trigonometric Equations. 

1. Find the value of A when cos A = \. 
We know that one value of A is 60° (§ 8). 

And since cos (- 60°)= cos 60° (§ 29), another value of A is - 60°. 
Again, by the principle of § 21, any multiple of 360° may be added to, 
or subtracted from, an angle, without altering its functions. 
Hence, other values of A are 

360° + 60°, 720° + 60°, -36G° + 60°, 360°- 60°, 720°-60°, -360°-60°, etc. 

It is evident from the above that the number of possible values of A 
is indefinitely great ; and that each is in the form 

n x 360° + 60°, or n x 360° - 60° ; 

where n is 0, or any positive or negative integer. 
Using the circular notation, we have 

A = n x27r±" = 2mr±-> 
o o 

2. Fino^ the value of A when tan A = -J- V3. 

We know that one value of A is 30° (§ 8) ; another is 180° + 30° (§ 27). 
Adding to, and subtracting from, these angles multiples of 360°, other 
values of A are 

360° + 30°, 540° + 30°, - 360° + 30°, - 180° + 30°, etc. 

It is evident from the above that all the values of A are given by the 
expression 

n X 180° + 30° ; 

where n is 0, or any positive or negative integer. 
Or, A = n v + ^ 

3. Find the value of A when sin A = £V2. 
One value of A is 45° (§ 7). 

And since sin (180° - 45°) = sin 45° (§ 33), another value of A is 
180° - 45°. 



38 PLANE TRIGONOMETRY. 

Adding to, and subtracting from, these angles multiples of 360°, other 
values of A are 

360° + 45°, 540° - 45°, - 360° + 45°, - 180° - 45°, etc. 

It is evident from the above that all the values of A are given by the 
expression 

rcxl80° + (-l)»45°; 

where n is 0, or any positive or negative integer. 

Or, A = n7r + (-l) n ^ 

It is evident that, to find the value of A in any equation of the above 
forms, we find any one of the values of A, and substitute it for A in the 
following expressions : 

If sin A is given, mr + (— T) n A. 
If cos A is given, 2 mr ± A. 
If tan A is given, mr + A. 

The rule for equations giving the value of cot A is the same as for 
tan .4; for sec A, the same as for cos A; and for esc A, the same as for 
sin A 

EXAMPLES. 

In each of the following find the value of A: 

4. tan^L=V3. 6. sin A=\. 8. cot^L=-l. 10. cot ^4=0. 

5. cos.4=— 1-V3. 7. sec4=V2. 9. esc A= -|V3. 11. sec A= -1. 

63. 1. Solve the equation cos 2 A = cos A. 

By (28), 2 cos 2 A — 1 = cos A, or 2 cos 2 A — cos A = 1. 



Solving this equation, cos A = 1:1= ^ 1 + 8 --= ±^ = 1 or - i. 

4 4 2 

If cos A=l, one value of A is 0° (§ 22), and ^1=2 mr (§ 62). 

If cos A = - 1, one value of A is 120° (§ 27), and ^1 = 2 mr ± ~ 

2. Solve the equation tan 2x= 6 tan #. 

One solution is evidently tan x = 0. 

In this case, one value of x is 0°, and x = mr (§ 62). 



MISCELLANEOUS THEOREMS. 



39 



Dividing (A) by 2 tan x, we have 
1 



1 — tan 2 x 



= 3, or 1 = 3 — 3 tan 2 x, or 3 tan 2 x = 2. 



Whence, 
Therefore, 



tair x 



/I 



-, or tana; = ±-y 



±W6. 



tan- 1 (± £V6) = ± tan" 1 (£V6). 



EXAMPLES. 

In each of the following find the value of x : 

3. sin x = sin 2 x. 7. cot 2 x + cot # = 0. 

4. sin 2 £ -f cos £ = 0. 8. tan (45°— x) -fcot (45° — aj) = 4. 

5. cos it* + cos 3 # = 0. 9. tan 3 x = 5 tan a?. 

6. tan 3 x + tan a; = 0. 10. cos x cot # = 1. 

rsj. T . ..• tt i r sinoj , tana 

64. Limiting Values of and — — 

x x 

m n , , ,. . . 7 ~ ^7 sin a; _ tan # • ' 

To ^wa the limiting values of the fractions — - — ana — - — when x is 

indefinitely decreased. 

Note. We suppose x to be expressed in circular measure (§ 52). 




Let OPXP' be a sector of a circle. 

Draw PT and P'T tangent to the arc at P and P', and join OT and 
PP'. 

By Geometry, PT= P'T. 

Then OT is perpendicular to PP' at its middle point M, and bisects 
the arc PP' at X. 

Let Z XOP = Z XOP' = a?. 

By Geometry, arc PP' > chord PP', and < PTP'. 

Whence, arc PX > PJf, and < PT. 

Therefore, *^*>^, and <*£ 

OP OP' OP 

Or by § 52, circ. meas. a; > sin x, and < tan x. 



40 PLANE TRIGONOMETRY. 

Representing the circular measure of x by x simply, and dividing 
through by sin x, we have 

x _ -, -, . tan x 1 
> 1, and <- or 



sin x sm x cos x 

Whence, < 1, and >cos#. 

x 

But when x is indefinitely decreased, cos x approaches the limit 1 

(§ 22). 

Hence, approaches the limit 1 when x is indefinitely decreased. 

tan x sin x sin x 1 
Again, = == x 



x x cos x x cos x 

But and approach the limit 1 when x is indefinitelv 

x cos x J 

decreased. 

Hence, approaches the limit 1 when x is indefinitely decreased. 

x 



LOGARITHMS. 41 

//$ -?3 - brty^V'' V - LOGARITHMS. 

65. Every positive number may be expressed, exactly or approxi- 
mately, as a power of 10. 

Thus, 100 = 10 2 ; 13 = 10 1113943 - ; etc. 

When thus expressed, the corresponding exponent is called its Loga- 
rithm to the Base 10. 

Thus, 2 is the logarithm of 100 to the base 10 ; a relation which is 
written log J0 100 = 2, or simply log 100 = 2. 

66. Logarithms of numbers to the base 10 are called Common Loga- 
rithms, and, collectively, form the Common System. 

They are the only ones used for numerical computations. 
' Any positive number, except unity, may be taken as the base of a 
system of logarithms ; thus, if of = m, where a and m are positive 
numbers, then x = log a m. 

Note. A negative number is not considered as having a logarithm. 



67. We have by Algebra, 






10° = 1, 




^Hr^ 


10 1 = 10, 




10-=^=. 01, 


10 2 = 100, 




10- 3 = i=.001,etc. 


Whence by the definition of § 


65, 




log 1 = 0, 




log .1=- 1 = 9-10, 


log 10 = 1, 




log.01 = -2 = 8-10, 


logl00 = 2, 




log.001=-3 = 7-10, etc 



Note. The second form for log.l, log .01, etc., is preferable in practice. If no 
base is expressed, the base 10 is understood. 

68. It is evident from § 67 that the logarithm of a number greater 
than 1 is positive, and the logarithm of a number between and 1 



69. If a number is not an exact power of 10, its common logarithm 
can only be expressed approximately. 

The integral part of the logarithm is called the characteristic, and the 
decimal part the mantissa. 



42 PLANE TRIGONOMETRY. 

For example, log 13 = 1.113943. 

In this case, the characteristic is 1, and the mantissa .113943. 

For reasons which wiH appear hereafter, only the mantissa of the 
logarithm is given in a table of logarithms of numbers; the characteristic- 
must be found by aid of the rules of §§ 70 and 71. 

70. It is evident from § 67 that the logarithm of a number between 

1 and 10 is equal to + a decimal ; 
10 and 100 is equal to 1 + a decimal ; 
100 and 1000 is equal to 2 + a decimal ; etc. 

Therefore, the characteristic of the logarithm of a number with one 
figure to the left of the decimal point, is ; with two figures to the left 
of the decimal point, is 1; with three figures to the left of the decimal 
point, is 2 ; etc. 

Hence, the characteristic of the logarithm of a number greater than 1 is 1 
less than the number of places to the left of the decimal point. 

For example, the characteristic of log 906328.51 is 5. 

71. In like manner, the logarithm of a number between 

1 and .1 is equal to 9 + a decimal — 10 ; 
.1 and .01 is equal to 8 + a decimal — 10 ; 
.01 and .001 is equal to 7 + a decimal — 10 ; etc. 

Therefore, the characteristic of the logarithm of a decimal with no 
ciphers between its decimal point and first significant figure, is 9, with 
— 10 after the mantissa; of a decimal with one cipher between its point 
and first significant figure is 8, with — 10 after the mantissa ; of a deci- 
mal with two ciphers between its point and first significant figure is 7, 
with — 10 after the mantissa ; etc. 

Hence, to find the characteristic of the logarithm of a number less than 1. 
subtract the number of ciphers between the decimal point and first significant 
figure from 9, writing — 10 after the mantissa. 

For example, the characteristic of log .007023 is 7, with — 10 written 
after the mantissa. 

Note. Some writers combine the two portions of the characteristic, and write 
the result as a negative characteristic before the mantissa. 

Thus, instead of 7.603658 - 10, the student will frequently find 3.603658, a minus 
sign being written over the characteristic to denote that it alone is negative, the 
mantissa being always positive. 



LOGARITHMS. 43 



PROPERTIES OF LOGARITHMS. 



72. In any system, the logarithm ofl is 0. 

For by Algebra, a = 1 ; whence by § 66, log a 1=0. 

73. In any system, the logarithm of the base is 1. 
For a 1 = a : whence, log„ a = 1. 



74. In any system whose base is greater than 1, the logarithm of is 
— x . 

For if a is greater than 1, <* _x = — - = — =0. 
Whence by § 66, log a = - oo . 

Note. No literal meaning can be attached to such a result as log a = — x> ; it 
must be interpreted as follows r 

If, iu any system whose base is greater than unity, a number approaches the limit 
0, its logarithm is negative, and increases without limit in absolute value. 



75. In any system, the logarithm of a product is equal to the sum of the 

logarithms of its factors. 
Assume the equations 

a x = m ) , ( x = log a m, 

- ; whence bv § 66, -■ 
a" = n ) " (y = hog a ?*. 

Multiplying the assumed equations, 

a x x a' J = mn, or a x+!/ =mii. 

Whence, \og a mn = x + y = log a m + log a u. 

In like manner, the theorem may be proved for the product of three 
or more factors. 

76. By aid of § 75, the logarithm of a composite number may be 
found when the logarithms of its factors are known. 

1. Given log 2 = .3010 and log 3 = .4771 ; find log 72. 
log72 = log(2 X 2x2x3x3) 

= log 2 + log 2 + log 2 + log 3 -f log 3 (§ 75) 

= 3 x log 2 + 2 x log 3 = .9030 + .9542 = 1.8572. 



44 



PLANE TRIGONOMETRY. 



EXAMPLES. 
Given log 2 = . 3010, log 3 = .4771. log 5 = .6990. log 7 = .8451, find : 

2. log 35. 6. log 126. 10. log 324. 14. log 2625. 

3. log 50. 7. log 196. 11. log 378. 15. log 6048. 

4. log 42. 8. log 245. 12. log 875. 16. log 12005. 

5. log 75. 9. log 210. 13. log 686. 17. log 15876. 

77. In any system, the logarithm of a fraction is equal to the logarithm 
of the numerator minus the logarithm of the denominator. 

Assume the equations 

" = m \l whence, .j* = } S. TO . 

a y = n ) (y = log a n. 

Dividing the assumed equations, 

a x m x _ v m 

— = —, or a x * = — 
a y n n 



Whence, 



log* - = x — hf= log a m — log a n. 

n 



78. 1. Given log 2 = .3010; find log 5. 

log 5 = log ^ = log 10 - log 2 (§ 77) = 1 - .3010 = .6990. 



EXAMPLES. 
Given log 2 == .3010, log 3 = .4771, log 7 = .8451, find 



2. 


log-- 


5. 


log 14f 


8. 


log i 


11. 


log 28* 


3. 


iog|- 


6. 


i 49 

log — 

& 27 


9. 


log 6|. 


12. 


, 200 

log 

° 9 


4. 


log 45. 


7. 


log 225. 


10. 


log 135. 


13. 


log 110* 



79. In any system, the logarithm of any power of a quantity is equal to 
the logarithm of the quantity multiplied by the exponent of the power. 

Assume the equation a x = m ; whence, x = log m. 
Raising both members of the assumed equation to the pth power, 
a p X _ m p . w h ence? \og a m p =px =p log a m. 



LOGARITHMS. 45 

80. In any system, the logarithm of any root of a quantity is equal to the 
logarithm of the quantity divided by the index of the root. 



For, log a Vm = log a (m r ) = - log a m (§ 79). 

r 



81. 1. Given log 2 = .3010 ; find log 2* 

log 2* = | x log 2 = | x .3010 = .5017. 

o o 

Note. To multiply a logarithm by a fraction, multiply first by the numerator, and 
divide the result by the denominator. 

2 Given log 3 = .4771 ; find log ^3. 

log J/3 = l2§3 = : 47n 05% 
8 8 



EXAMPLES. 
Given log 2 = .3010, log 3 = .4771, log 7 = .8451, find : 

3. log3 7 . 6. log28 6 . 9. log v2. 12. log a/525. 

4. log 5*. 7. log 18*. 10. log ^5. 13. log ^294. 

5. log 7*. 8. log 96*. 11. log ^/7. 14. log ^216. 

15. Find log (2* x 3*). 

By § 75, log (23- x 3*) = log 2* + log 3* = $ log 2 + f log 3 
= .1003 + .5964 = .6967. 

Find the values of the following : 

16. log^|- 18. Iog(2*xl0*). 20. log^l- 22. log- 



3^ 



17. logg)* 



</3 V24 

19. log 7^2. 21. log 51 23. log^. 

7* 5* 



82. To prove the relation 

, log a m 

log 6 m = — ^— • 

loga & 

Assume the equations 



aX = m \. whence, i aj = l°g. m » 
b y = m ,) \ y = log* m. 



46 PLANE TRIGONOMETRY. 

From the assumed equations, a x = b v . 

X 

Taking the y%\\ root of both members, a y = 6. 
Therefore, log a b =-, or y = 



That is, log 6 m = 



log> 
log m 



log tt 6 



83. To prove the relation 

log 6 a x log a 6 = 1 
Putting m = a in the result of § 82, we have 

logsa = |^ = 1 (J73). 

log a b log a b v ' 

Whence, log 6 a x log a 6 = 1. 

84. In the common system, the mantissce of the logarithms of numbers 
having the same sequence of figures are equal. 

Suppose, for example, that log 3.053 = .484727. 

Then, log 305.3 = log (100 x 3.053) = log 100 + log 3.053 

= 2 + .484727 = 2.484727 ; 
log .03053 = log (.01 x 3.053) = log .01 + log 3.053 

= 8 - 10 + .484727 = 8.484727 - 10 ; etc 

It is evident from the above that, if a number be multiplied or divided 
by any integral power of 10, producing another number with the same 
sequence of figures, the mantissse of their logarithms will be equal. 

The reason will now be seen for the statement made in § 69, that only 
the mantissse are given in a table of logarithms of numbers. 

For, to find, the logarithm of any number, we have only to take from 
the table the mantissa corresponding to its sequence of figures, and the 
characteristic may then be prefixed in accordance with the rules of §§ 70 
or 71. 

Thus, if log 3.053 = .484727. then 

log 30.53 = 1.484727. log .3053 = 9.484727 - 10, 

log 305.3 = 2.484727. log .03053 = 8.484727 - 10. 

log 3053. = 3.484727, log .003053 = 7.484727 - 10, etc. 

This property is only enjoyed by the common system of logarithms, 
and constitutes its superiority over others for the purposes of numerical 
computation. 



LOGARITHMS. 47 

85. 1. Given log 2 = .3010, log 3 = .4771 ; find log .00432. 
We have, log 432 = log (2 4 x 3 3 ) = 4 log 2 + 3 log 3 = 2.6353. 
Then by § 84, the mantissa of the result is .6353. 
Whence by § 71, log .00432 = 7.6353 - 10. 

EXAMPLES. 
Given log 2 = .3010, log 3 = .4771, log 7 = .8451, find: 

2. log 3.6. 6. log .00343. 10. log .1944. 

3. log 11.2. 7. log 2880. 11. log 202.5. 

4. log .84. 8. log .0392. 12. log a/6\4. 

5. log .098. 9. log .000405. 13. log (14.7)1 

USE OF THE TABLE OF LOGARITHMS OF NUMBERS. 

(For directions as to the use of the Table of Logarithms of Numbers, 
see pages iii to v of the Introduction to the Author's Six Place Logarith- 
mic Tables.) 

EXAMPLES. 

86. Find the logarithms of the following numbers : 

1. .053. 5. 336.908. 9. .001030746. 

2. 51.8. 6. .000602851. 10. .00000876092. 

3. .2956. 7. 65000.63. 11. 730407.8. 

4. 1.0274. 8. 9122.55. 12. .0000436927. 

Find the numbers corresponding to the following logarithms : 

13. 1.880814. 17. 8.044891-10. 21. 3.990191. 

14. 9.470410-10. 18. 2.270293. 22. 5.670180. 

15. 0.820204. 19. 7.350064-10. 23. 6.535003-10. 

16. 4.745126. 20. 5.000027-10. 24. 4.115658-10. 

APPLICATIONS. 

87. The approximate value of an arithmetical quantity, in which the 
operations indicated involve only multiplication, division, involution, or 
evolution, may be conveniently found by logarithms. 

The utility of the process consists in the fact that addition takes the 
place of multiplication, subtraction of division, multiplication of invo- 
lution, and division of evolution. 

Note. In computations with six-place logarithms, the results cannot usually be 
depended upon to more than six significant figures. 



48 PLANE TRIGONOMETRY. 

88. 1. Find the value of .0631 x 7.208 x .51272. 

By § 75, log (.0631 x 7.208 x .51272) = log .0631 + log 7.208 + log .51272. 

log .0631= 8.800029-10 
log 7.208= 0.857815 
log .51272= 9.709880-10 

Adding, log of result = 19.367724 - 20 = 9.367724 - 10. (See Note 1.) 

Number corresponding to 9.367724 - 10 = .233197. 

Note 1. If the sum is a negative logarithm, it should be written in such a form 
that the negative portion of the characteristic may be — 10. 
Thus, 19.367724 - 20 is written in the form 9.367724 - 10. 

2. Find the value of 336 - 852 . 
7980.04 

By § 77, log ilri = log 336,852 " log 798a04 - 

log 336.852 = 12.527439 - 10 (See Note 2.) 
log 7980.04= 3.902005 



Subtracting, log of result = 8.625434 — 10 

Number corresponding = .0422118. 

Note 2. To subtract a greater logarithm from a less, or to subtract a negative 
logarithm from a positive, increase the characteristic of the minuend by 10, writing 
— 10 after the mantissa to compensate. 

Thus, to subtract 3.902005 from 2.527439, write the minuend in the form 
12.527439 - 10 ; subtracting 3.902005 from this, the result is 8.625434 - 10. 

3. Find the value of (.0980937) 5 . 
By § 79, log (.0980937) 5 .= 5 x log .0980937. 

log .0980937 = 8.991641 - 10 



44.958205 - 50 = 4.958205 - 10. (See Note 1.) 
Number corresponding = .0000090825. 



4. Find the value of V.035063. 



By § 80, log V.035063 = \ log .035063. 

log .035063 = 8.544849 - 10 

3 )28.544849 - 30 (See Note 3.) 
9.514950 - 10 

Number corresponding = .327303. 



LOGARITHMS. 49 

Note 3. To divide a negative logarithm, write it in such a form that the negative 
portion of the characteristic may be exactly divisible by the divisor, with — 10 as the 
quotient. 

Thus, to divide 8.544849 — 10 by 3, we write the logarithm in the form 
•28.544849 - 30; dividing this by 3, the quotient is 9.514950 - 10. 

89. Arithmetical Complement. 

The Arithmetical Complement of the logarithm of a number, or, briefly, 
the Cologarithm of the number, is the logarithm of the reciprocal of that 
number. 

Thus, colog 409 = log -i- = log 1 - log 409. 

log 1 = 10. - 10 (Note 2, § 88.) 

log 409= 2.611723 

Then, colog 409= 7.388277-10. 

Again, colog .067 = log—-- = log 1 — log .067. 

.06/ 

logl = 10. -10 

log .067= 8.826075-10 

Then, colog .067= 1.173925. 

It follows from the above that the cologarithm of a number may be 
found by subtracting its logarithm from 10 — 10. 

Note. The cologarithm may be obtained by subtracting the last significant 
figure of the logarithm from 10, and each of the others from 9, — 10 being written 
after the result in the case of a positive logarithm. 



90. Example. Find the value of ' 



1.384 



8.709 x .0946 

l ° s s,!ofLe = l0 < 51 - 384 x 8J09 ' iob) 

= log 51.384 + colog 8.709 + colog .0946. 
log 51.384 = 1.710828 
colog 8.709 = 9.060032 - 10 
colog .0946 = 1.024109 



1.794969 = log 62.369. 

It is evident from the above example that the logarithm of a fraction 
whose terms are composed of factors may be found by the following rule : 

Add together the logarithms of the factors of the numerator, and the 
cologarithms of the factors of the denominator. 



50 



PLANE TRIGONOMETRY. 



Note. The value of the above fraction may be found without using cologarithms, 
by the following formula : 

51.384 



log 



log 51.384 - log (8.709 x .0946) 



8.709 x .0940 

= log 51.384 -(log 8.709 + log .0946). 

The advantage in the use of cologarithms is that the written work of computation 
is exhibited in a more compact form. 



EXAMPLES. 

Note. A negative quantity has no common logarithm (§ 66, Note). 

If such quantities occur in computation, they should be treated as if they were 
positive, and the sign of the result determined irrespective of the logarithmic work. 

Thus, in Ex. 2, § 91, the value of 84.759 x (- 2280.76) is obtained by finding the 
value of 84.759 x 2280.76, and putting a negative sign before the result. See also 
Ex. 29. 



91. Find by logarithms the values of the following : 



3.1425 x 603.93. 
84.759 x (-2280.76). 

4867.2 6 1.05478 

765.16* ' 34.9564* 

3.89612 x .6946 



10. 



4694.9 x .00454 

715 x (- .024158) 
(-.5157) x 1420.63' 



3. 

4. 

7. 
11. 
12. 



(-4.39182) x (-.0703968). 
.936537 x .00117854. 

2.7085 a - .000680239 



.0868097 



8. 



.00512643 



13. 
14. 
15. 
16 
17. 



(7.7954) 4 . 

(.83287) 7 . 

( - 25.1437) 3 . 

(•01)". 

( - 964.38)*. 

28. Find the value of 



18. (.0951293)*. 

19. (.000105936) 

20. VS. 

V2. 
V=6. 



21. 
22. 

2V5 
3* ' 



(- .870284) x 3.73 
(-.06585) x (-42.317) 
.082136 x (- 73.39) 

.838 x 2808.72 

23. ^100. 

24. V19916. 
25 
26 
27 



V.0725628. 
V.002613874. 

V- 



000951735. 



By § 90, 

log ^y~^ = log 2 + log V5 + colog 3* = log 2 + J log o + -jf colog 3. 
3* 

log 2 = .301030 
log 5= .698970; divide by 3 = .232990 

colog 3 = 9.522879 - 10 ; multiply by f = 9.602399 - 10 



.136419 = log 1.36905. 



LOGARITHMS. 51 



29 . x ~-'°f i 2 !f 6 - 

96183 



v* 



log i/ - 032956 = i log -° 32 ^ = i (log .032956 - log 7.96183). 
& \ 7.96183 3 & 7.96183 3V & b ; 



7.96183 * ° 7.96183 

log .032956 = 8.517934 - 10 
log 7.96183 = 0.901013 



3 )27.616921 - 30 
-10 

Result l <,-■-■' 1 



9.205640 - 10 = log .160561. 



Find the values of the following 



30. 


. 4 _2 

4 3 x < 3 . 


31. 


8»* 


32. 


10/79 


33 


(.001)5- 




^7 


34 


V^08 




(-10)* 



35. 



36. yj 



4400 \* 40. v"3 x S/5 x Vl. 

ftQ97 7 J ' 3 

'76.1 x .05929Y* 



41. 



/27085; \ 1.307;- 

940 



38. 



5^_ 42. 

=3 
-^1000 43. 



v 



.438 



37. • ^ \ 31.4 x .4146 

S/-.1 



</.000965782 



V .00497836 



(-.6)* 

5 /7 44. - (-256929)' 

(- .834574)' 



39 s/i-% 



45. (25.4673) 10 x (- .052)'-. 5Q (.5732)- 

46. ^5106.526 x .000031093. " 8693.84 x -^.033074* 

47. (837.48 x .00943246)'. ( - .000191 62)^ x V68l8 

48. (4.867184)* x (.176437)* * -.2755653 
49 V3.9285 x-y 65.4775 52 -J/.052866 



V721.329 VC374 x ^.007835912 

EXPONENTIAL EQUATIONS. 

92. An Exponential Equation is an equation of the form of = b. 
To solve an equation of this form, take the logarithms of both mem- 
bers. 

1. Given 31* = 23 ; find the value of x. 
Taking the logarithms of both members, 

log (31*) = log 23. 
Whence by § 79, x log 31 = log 23. 

Then, ^ = jgg|3 = 1.361728 9130+> 

' log 31 1.491362 



52 PLANE TRIGONOMETRY. 

2. Given .2* = 3 ; find the value of x. 
Taking the logarithms of both members, 

x log .2 = log 3. 

m log 3 .477121 .477121 » QOr , 

^\ hence, x = — ^ — = ■. = = — .6826 -K 

log .2 9.301030-10 -.698970 

EXAMPLES. 

Solve the following equations : 

3. 332.9* = 5.178. 5. .0158* = .0082958. 7. a* = &*<* 

4. .4162* = 6.724. 6. 5.3364- = .744. 8. m 2 a* = n*. 

9. 6 2 "- 8 = . 0277778. 10. .7* 2+4l = .16807. 

93. 1. Find the logarithm of .3 to the base 7. 

By § 82, lo g7 .3 = 1^3 = 9.477121 - M> = - -522879 = _ 618J + 
y ' 6 ' log 10 7 .845098 .845098 

EXAMPLES. 

Find the values of the following : 

2. log 2 13. 4. log. 74 6.2. 6. logcu .362. 

3. log 3 .9. 5. log. 48 .087. 7. lo gG5 4.3. 

Examples like the above may be solved by inspection if the number 
can be expressed as an exact power of the base. 

8. Find the logarithm of 128 to the base 16. 

Let log 16 128 = x ; then by § 66, 16* = 128. 

That is, (2 4 )* = 2 7 , or 2^ = 2 7 . 

Whence by inspection, 4 x = 7 ; and x = log I6 128 = — 

4 

9. Find the logarithm of 81 to the base 3. 

10. Find the logarithm of 32 to the base 8. 

11. Find the logarithm of -J- to the base 27. 



12. Find the logarithm of ¥ * ¥ to the base -gL-. 



LOGARITHMS. 53 

EXAMPLES IN THE USE OF TRIGONOMETRIC TABLES. 

(For directions, see pages v to xi of the Introduction to the Author's 
Six Place Logarithmic Tables.) 

94. Table of Logarithmic Sines, Cosines, etc. 

Find the values of the following : 

1. log sin 12° 48 '52". 4. log cot 53° 42 '9". 7. log cot 26° 30' 14''. 

2. log tan 67° 13 '27". 5. log cos 79° 54' 35". 8. log sec 45° 26 '38". 

3. log cos 31° 5 '43". 6. log tan 8° 17' 21". 9. log esc 84° 9 '50'. 

Find the angles corresponding in the following: 

10. log sin = 9.934232 - 10. 14. log tan = 9.1843G7 - 10. 

11. log cos = 9.923569 - 10. 15. log cot = 9.404692 - 10. 

12. log tan = 0.806571. 16. log sec = 0.188783. 

13. log cot = 0.282956. 17. log esc = 0.400314. 

95. Table of Natural Sines, Cosines, etc. 
Find the values of the following : 

1. sin43°17'35". 3. cos 86° 21 '46". 5. sin67°9'54". 

2. cot75°50'19". 4. tan34°48'23". 6. cos29°35'8". 

Find 'the angles corresponding in the following : 
7. tan = 1.2622. 8. cos = .96376. 9. sin = .91527. 10. cot = 1.7927 

96. Auxiliary Table for Small Angles. 

Find the values of the following : 
1. log sin 1° 14' 53". 2. log tan 3° 42 '8". 3. log cot 2° 26' 35 '. 

Find the angles corresponding in the following : 
4. log sin = 8.233459 - 10. 5. log tan = 7.859872 - 10. 

6. log cot = 1.546267. 



54 



PLANE TRIGONOMETRY. 



VI. SOLUTION OF RIGHT TRIANGLES. 

97. The elements of a triangle are its three sides and its three angles. 

We know by Geometry that a triangle is, in general, completely deter- 
mined when three of its elements, are known, provided one of them is a 
side. 

The solution of a triangle is the process of computing the unknown 
from the given elements. 



98. To solve a right triangle, two elements must be given in addition 
to the right angle, one of which must be a side. 

The various cases which can occur may all be solved by aid of the 
following formulae : 



. . a 
sin A = — 

c 



sin B 




tan A = -- 

o 



tan B = — 

a 



99. Case I. When the given elements are a side and an angle. 

The proper formula for computing either of the remaining sides may 
be found by the following rule: 

Take that function of the angle which involves the given side and the 
required side. 

1. Given c = 68, B = 21° 42' 39"; find a and b. 
In this case the formulae to be used are 

b 



AY hence, 



cos B = -, and sin B 

c c 

a = c cos B, and b = c sin B. 



(A) 



Solution by Natural Functions. 

a -68 x cos 21° 42' 39" = 68 x .92906 = 63.176. . 
b = 68 x sin 21° 42' 39" = 6H x .36993 = 25.155. 



SOLUTION OF RTGHT TRIANGLES. 55 

Saint ioi) by Logarithms. 

Taking the logarithms of both members, in formulae (A), 

log a = log c + log cos B, and log b = log c -f log sin B. 

log c = 1.832509 log c = 1.832509 

log cos B = 9.968045 - 10 log sin £ = 9.568111 - 10 



log 6 = 


■. 1.400620 


b = 


: 25.1547. 


ind & and 


c. 


4 = 1 

c 




a 





log a = 1.800554 
a = 63.1762. 

2. Given a = .235867, .4 = 67° 9' 23 

In this case, tan A = -, and sin ^4 

6 

"Whence, 6 = , and c = . 

tan A smJ. 

By logarithms, log b = log a — log tan A, and log c = log a — log sin A. 

log a = 9.372667 - 10 log a = 9.372667 - 10 

log tan A = 0.375452 log sin A = 9.964527 - 10 

log b = 8.997215 - 10 log c = 9.408140 - 10 

b = .0993607. c = .255941. 

100. Case II. When both the given elements are sides. 

First calculate one of the angles by aid of either formula involving 
the given elements, and then compute the remaining side by the rule of 
Case I. 

Example. Given b = .15124, c = .30807 ; find A and a. 

We first find A by the formula cos A = -, and then find a by the 

a c 

formula sin' A = -, or a = c sin A. 

c 

By logarithms, log cos A = log b — log c, and log a = log c + log sin A. 

log b = 9.179667 - 10 log c = 9.488650 - 10 

log c = 9.488650 - 10 losr sin A = 9.940118 - 10 



log cos A = 9.691017 - 10 log a = 9.428768 - 10 



A = 60° 35' 54.4". a = .268391. 



101. In the Trigonometric solution of any example under Case II., 
it is necessary to first find one of the angles, and the remaining side may 
then be calculated. 

It is possible, however, to compute the third side directly, without first 
finding the angle, by Geometry. 



56 PLANE TRIGONOMETRY. 

Thus, in the example of § 100, we have, by Geometry, ar + b 2 = c 2 . 



Whence, a = v'er — b 2 = V (c + 6) (c — 6). 

By logarithms, log a = i [log (c + 6) + log (c — &)]. 

c + 6 = .45931 ; log = 9.662106 - 10 
c - 6 = .15683 ; log = 9.195429 - 10 



2 )18.857535-20 
log a = 9.428768 -10 
a = .268391, as before. 



If the given sides are a and b, the expression for c is Va 2 + b 2 , which 
is not adapted to logarithmic computation. 

In such a case, it is usually shorter to proceed as in § 100. 

^J^ l|l ' EXAMPLES. 

Note. In those examples of the following set in which the given sides are num- 
l»crs of not more than three significant figures, and the operations indicated involve 
only multiplication, it is usually shorter to employ Natural Functions. 

In such a case, the results cannot be depended upon to more than five significant 
figures ; while in the solutions by logarithms, they can be depended upon to six signifi- 
cant figures. 

102. Solve the following right triangles : 

1. Given A = 15°, c = 7. 9. Given A = 9°, 6 = 937. 

2. Given B = 67°, a = 5. 10. Given a = 3.414, b = 2.875. 

3. (riven B = 50°, b = 20. 11. Given A = 84° 16', a = .0033503. 

4. Given a = .35, c = .62. 12. Given A = 46° 23', c = 5278.6. 

5. Given a = 273, b = 418. 13. Given a = 529.3, c = 902.7. 

6. Given ^4 = 38°, a = 8.09. 14. Given B = 23° 9', 6 = 75.48. 

7. Given B = 75°, c = .014. 15. Given A = 72° 52'. b = 6306. 

8. Given b = 58.6, c = 76.3. 16. Given B= 18° 38', c = 2.5432. 

17. Given a = .0001689, b = .0004761. 

18. Given A = 31° 45', a = 48.0408. 

19. Given 6 = 617.57, c = 729.59. 

20. Given B = 82° 6' 18", a = 89.32. 

21. Given A = 55° 43' 29", c = 41518. 

22. Given B= 31° 47' 7". a = 7.23246. 

23. Given a = 99.464, c = 156.819. 



SOLUTION OF 1UGHT TRIANGLES. 57 

24. Given A = 43° 21' 36", b = .00261751. 

25. Given B = 79° 14' 31", 6 = 84218.5. 

26. Given B = 67° 39' 53", e = 9537514. 

27. Given 6 = 5789.72, c = 24916.45. 

28. Given A = 26° 12' 24", c = 469422.7. 

29. Given j5 = 14° 55' 42", b = .1353371. 

30. Given a = 672.3853, 6 = 384.5038. 

Solve the following isosceles triangles, in which A and B are the equal 
angles, and a, b, and c the sides opposite the angles A, B, and C, respec- 
tively : 



31. 


Given A = 68° 57', 


6 = 


= 350.94. 


32. 


Given B = 27° 8', 


c = 


= 3.0892. 


33. 


Given C= 84° 47', 


b = 


= 91032.7. 


34. 


Given a = 79.2434, 


c = 


= 106.6362, 


35. 


Given .4 = 35° 19' 47" 


, c = 


= .56235. 



36. Given C= 151° 28' 52", c = 9547.12. 

37. A regular pentagon is inscribed in a circle whose diameter is 35. 
Find the length of its side. 

38. At a distance of 105 ft. from the base of a tower, the angle of 
elevation of its top is observed to be 38° 25'. Find its height. 

39. What is the angle of elevation of the sun when a tower whose 
height is 103.74 ft. casts a shadow 167.38 ft. in length ? 

40. If the diameter of a circle is 32689, find the angle at the centre 
subtended by an arc whose chord is 10273. 

41. If the diameter of the earth is 7912 miles, what is the distance of 
the remotest point of the surface visible from the summit of a mountain 
l-\ miles in height ? 

42. Find the length of the diagonal of a regular pentagon whose side 
is 6.3257. 

43. Find the angle of elevation of a mountain-slope which rises 238 ft. 
in a horizontal distance of one-eighth of a mile. 

44. From the top of a lighthouse, 146 ft. above the sea, the angle of 
depression of a buoy is observed to be 21° 46'. Find the horizontal dis- 
tance of the buoy. 



58 PLANE TRIGONOMETRY. 

45. If a pole casts a shadow which is two-thirds its own length, what 
is the angle of elevation of the sun ? 

46. A vessel is sailing due east at the rate of 7.8 miles an hour. A 
headland is observed to bear due north at 10.37 a.m., and 33° west of 
north at 12.43 p.m. Find the distance of the headland from each point of 
observation. 

47. If a chord whose length is 41.368 subtends an arc of 145° 37', what 
is the radius of the circle ? 

48. The length of the side of a regular octagon is 12. Find the radii 
of the inscribed and circumscribed- circles. 

49. How far from the foot of a flagpole 110 ft. high must an observer 
stand, so that the angle of elevation of the top of the pole may be 12° ? 

50. If the diagonal of a regular pentagon is 32.835, what is the radius 
of the circumscribed circle ? 

51. From the top of a tower, the angle of depression of the extremity 
of a horizontal base line, 1250 ft. in length measured from the foot of the 
tower, is observed to be 18° 36' 29". Find the height of the tower. 

52. If the radius of a circle is 723.294, what is the length of a chord 
which subtends an arc of 35° 13'? 

53. A regular hexagon is circumscribed about a circle whose diameter 
is 18. Find the length of its side. 

54. From the top of a lighthouse 200 ft. above the sea. the angles of 
depression of two boats in line with the lighthouse are observed to be 14° 
and 32°, respectively. Find the distance between the boats. 

55. A vessel is sailing due east at a uniform rate of speed. At 7 a.m.. 
a lighthouse is observed bearing due north, 10.326 miles distant ; and at 
7.30 a.m. it bears 18° 13' west of north. Find the rate of sailing of the 
vessel, and the bearing of the lighthouse at 10 a.m. 



103. Care must be taken to use the Auxiliary Table for Small Angles 
in rinding the logarithmic functions of angles between 0° and 5°, or between 
85° and 90°, or the angles corresponding in the same cases. 

This provides for every case which can arise in solving right triangles, 
except in looking out the angle corresponding to a logarithmic sine when 
between 85° and 90°, or a logarithmic cosine when between 0° and 5°. 

AVe will now derive a formula for right triangles by aid of which, 
when b and c are given, the angle A may be determined with accuracy 
if it is between 85° and 90°. 





SOLUTION OF RIGHT TRIAXGI 


ES. 


By § 98, 


cos A = — 

c 




Then by (31), 


2 sin 2 \ A = 1 — cos A = 1 - 


6 c-6 

c c 


Therefore, 


™M=-\fer 





59 



In like manner, sin \B- 



m lc 



These formulae involve the half-angles; hence, if the angle itself is 
between 85° and 90°, its half is between 42° 30' and 45°, and the cor- 
rection in seconds may in that case be found from the table with sufficient 
precision. 

An angle between 0° and 5° may always be avoided in solving a right 
triangle by working with the other acute angle. 

104. 1. Given b = 1.08249, c = 1.08261 ; find the angles. 
Here A is near to 0°, and B is near to 90°, as may be determined by 
inspection. 

We then proceed to find B by the formula of § 103. 

For this purpose, we must first find a, which may be done as in § 101. 

c + b = 2.1651 ; log = 0.335478 
c-b = .00012 ; log = 6.079181 - 10 



2 )16.414659-20 
loga = 8.207330 -10 

Whence a = .0161187. 

Now to find B, we use the formula sin } I B=\j— 

By logarithms, log sin \ B = i [log (c — a) — log 2 c] . 

c-a = 1.0664913 ; log = 0.027957 
2 c = 2.16522 ; log = 0.335502 



2 )19.692455-20 
log sin i B = 9.846228 - 10 

Whence, £ B = 44° 34' 24.7 ". 

Then, B = 89° 8' 49.4", and A = 90° - B = 0° 51' 10.6". 

If b is small compared with c, then A is near to 90°, and should be 
calculated directly by aid of the formula of § 103. 



60 PLANE TRIGONOMETRY. 

EXAMPLES. 
In each of the following right triangles find the angles ; 

2. Given a = .0128, c = 152.337. 

3. Given b = 5.81006, c = 5.81039. 

4. Given c = 11527.2, b = 1.32. 

5. Given a = .77, c = 98276.4. 

6. Given a = 42.0098, c = 42.0103. 

FORMULAE FOR THE AREA OF A RIGHT TRIANGLE. 

105. Case I. Given the hypotenuse and an acute angle. 




Denoting the area by K, we have by Geometry, 

2K=ab. 
But by § 4, a = c sin A, and b = c cos A. 

Whence, 2 K = c 2 sin ^4 cos ^4 = \ c 2 sin 2 A, by (25). 
Then, 4 K= c 2 sin 2 A (38) 

In like manner, 4 K = c 2 sin 2 i?. (39) 

Case II. Given an angle and its opposite side. 

By § 4, b = a cot A. 

Whence, 2 K— a, x a cot A = a 2 cot A. (40) 

In like manner, 2K=^b 2 cot 5. (41 ) 

Case III. Given an angle and its adjacent side. 

By §4, b = atsmB. 

Whence, 2K=a x a tan B = a 2 tan B. (42) 

In like manner, 2K=b 2 tan^l. (43) 



SOLUTION OF RIGHT TRIANGLES. 61 

Case IV. Given the hypotenuse and another side. 
By Geometry, b 2 = c 2 — a 2 . 



Whence, 2 K= ab — aVc 2 — a 2 = a V(c + a)(c — a). (44) 



In like manner, 2 K= &V(c + b) (c — 6). (45) 

Case V. Given the two sides about the right angle. 

In this case, 2 K= ab. (46) 

EXAMPLES. 

106. 1. Given c = 10.3572, B = 74° 57' 14"; find the area. 

By (39), 4/f=c 2 sin2 J B. 

Whence, log (4 iT) = 2 log c + log sin 2 jB. 

' log c - 1.015242 ; multiply by 2 = 2.030484 

2B = 149° 54' 28" ; log sin = 9.700178 - 10 

log (4 K) = 1.730662 

4 K= 53.7851 

Dividing by 4, K= 13.4463. 

Note. To find log sin 149° 54' 28", take either log cos 59° 54' 28", or log sin 30° 
5' 32". (See Introduction to Tables, page viii.) 

Find the areas of the following right angles : 

2. Given A -19° 36', a = 2.2178. 4. Given a = 149.417, b = 76.292. 

3. Given £ = 24° 7 '48", a = .8213. 5. Given b = .305694, c=. 660156. 

6. Given A = 30° 56' 19", c = 192.035. 

7. Given .4=78° 42' 53", 6 = .0520281. 

8. Given a = .932368, c = 4.786723. 

9. Given B = 72° 18' 27", c = 27.28338. 
10. Given B = 49° 25' 34", b = .3375494. 



62 



PLANE TRIGONOMETRY. 



VII. GENERAL PROPERTIES OF TRIANGLES. 

107. In any triangle, the sides are proportional to the sines of their 
opposite angles. 

I. To prove a :b = sin A : sin B. (47) 

C 





Fig. 1. 

There will be two cases, according as the angles A and B are both 
acute (Fig. 1), or one of them obtuse (Fig. 2). 
In each case, draw CD perpendicular to AB. 
Then in each figure, CD = b sin A (§ 4). 

Also in Fig. 1, CD = a sin B. 

And in Fig. 2, CD = a sin CBD 

= a sin (180° - B) = a sin B (§ 33). 
Then in either case, b sin A — a sin B. 



Whence by the theory of proportion, 

a : b = sin A : sin B. 

In like manner, b : c = sin B : sin C, 

and c : a ■ = sin C : sin A. 



(48) 

(49, 



108. //* any triangle, the sum of any two sides is to their difference as 
the tangent of half the sum of the opposite angles is to the tangent of half their 
difference. 

By (47), a:b — sin A : sin B. 

Whence by composition and division, 

a -h b : a — b = sin A + sin B : sin A — sin B. 
a + b sin A + sin B 



Or, 
But 



a — b sin A — sin B 

sin A + sin 5 _ tan i (J. + 5) 
sin A — sin 1? — tan ± (A — B) 



. »y (2D- 



GENERAL PROPERTIES OF TRIANGLES. 63 

^ hence - E = ^ \ a — m' ( 50 ) 



In like manner, = = - f^— -^-, (51) 



a — b 


tan -J 


(-1- 


B) 


b + c 


tani 


(B + 


0) 


b-c 


tani 


(B- 


cy 


c + a 


tan-J 


(G + A) 



aml . —01 = ^1(0 -A) (52) 

109. In any triangle, the square of any side is equal to the sum of the 
squares of the other two sides, minus twice their product into the cosine of 
their included angle. 

I. To prove a 2 = b 2 -f c 2 - 2 be cos A. -(53) 

'Case I. When the included angle A is acute. (Figures of § 107.) 

There will be two cases, according as the angle B is acute (Fig. 1), or 
obtuse (Fig. 2). 

Then in Fig. 1, BD = c - AD, and in Fig. 2, BD = AD - c. 
Squaring, we have in either case, 

BD' = AD 2 -f c 2 - 2 c x AD. 

Adding CD' to both members, 

BD 2 + CD 2 = AD 2 + CD 2 + c 2 - 2 c x ^4Z). 

But 5Z> 2 + Cl? - a 2 , and ZZ) 2 + OZ) 2 = b 2 . 

Also, by § 4, ^LD = & cos A. 

Whence, - a 2 = b 2 + c 2 — 2 be cos A 

Case II. Wlien the included angle A is obtuse. 

C 



B c A D 

Draw CD perpendicular to AB. 

We have BD = AD + c. 

Squaring, and adding CD to both members, 

BD 2 +CD 2 = AD 2 + CD 2 + <? + 2cxAD. 



(34 PLANE TRIGONOMETRY. 

But BD 2 + W=a 2 , and AL? + CD* = b\ 

And by § 4, AD = b cos CAD = b cos (180° - A) = - b cos A (§ 33). 
Whence, a 2 = 6 2 + c 2 — 2 be cos A 

In like manner, b 2 = c 2 + a 2 — 2 ca cos 5, (54) 

and c 2 = a 2 + 6 2 -2a6cosC. (55) 

110. To express the cosines of the angles of a triangle in terms of the 
sides of the tnangle. 

By (53), a 2 = ft 2 + c 2 - 2 be cos A 

Transposing, 2 6c cos ^1 = b 2 -\- c 2 — a 2 . 

h 2 4- {$ — a 2 

Whence, cos A = ^tf — - • (56 ) 

Z be 

c 2 4- a 2 b 2 

In like manner, cos B = — 3— , (57) 



and cos C = ~*~ ~ • (58 ) 

2ab v 



111. To express the sines, cosines, and tangents of the half-angles of a 
triangle in terms of the sides of the triangle. 

By (56), l-cos^^l- ^ + c2 - a ^ a2 - 62 + 26c - c2 . 

J K " 2 be 2 be 

Whence, by (31), 2 sin 2 ±A = a " ~{ b ~ C ^ - 

2 be 

Or sin 2 1 4 - (« - & + c)(a + 6- c) 

Or, sin Y A- — 

Denoting the sum of the sides, a + b + c, by 2 s, we have 

a - b + c = (a + b + c) -2 6 = 2s — 2b = 2(s -6), 
and a + & — c = (a + & + c) — 2c = 2s — 2c = 2(s — c). 

Whence, sin 2 1 .4 = 4('-&)('-<0 , 

46c 



cos 


4 = 


_ 6 2 + c 2 - 


•a 2 


2 6c 






5 = 


c 2 + a 2 - 


6 2 




2ca 


9 


POS 


n- 


_ a 2 + 6 2 - 


■c 2 



Or, sin ^ 4=N/ '(^(iH5. (59) 



GENERAL PROPERTIES OF TRIANGLES. 65 



In like manner, sm-i-B=^ & \ (60) 



and sinj0 =V (S — ^'-- 6) - (61) 

Again, by (56), 1 + cos^ = 1 + * +f~ a ' = y + 2 ^ + * ~ < 



Whence, by (32), 2 cos 2 \A = (fe + f? ~ "' • 

£ oc 



cog2 1 ^ = (& + c + a)_(6 + c-a^ 

But, b +c + a = 2 s; 

and 6 -f- c — a — (6 + c + a) — 2 a = 2 (s — a). 

Whence, cos 2 £ A = iiiizi^. 

2 4 6c 



Or, cos i .4 = ^ g ~ a ) . (G2) 



In like manner, cos ^ 5 = \-^- K (63) 

* ca 



and 

Dividing (59) by (62), we have, by (4), 



cos JC=^p* & (64) 



tan i ^ /(*-&)(*-c) / fte = J(«-&)(«-c). ( } 

\ 0€ \ s ( s -a) \ a(*-a) V ; 



In like manner, tan -J- .B =\/^— ^ ^ (66) 



«(* - b) 



and tan £ C = J ('-«) (»-») . (67) 

^ s(s — c) 

Note. Since each angle of a triangle is less than 180°, its half is less than 90° ; 
hence, the positive sign must be taken before the radical in each formula of § 111. 



FORMULA FOR THE AREA OF AN OBLIQUE TRIANGLE. 

112. Case I. Given two sides and their included angle. 
I. When the given parts are b, c, and A. 



K6 



PLANE TRIGONOMETRY. 





Fig. 2. 



There will be two cases, according as A is acute (Fig. 1), or obtuse 
(Fig. 2). 

In each case, draw CD perpendicular to AB. 

Then denoting the area by K, we have by Geometry, 

2K=cx CD. 

CD = b sin A (§ 4). 

CD = b sin CAD 

= b sin (180° - A) = b sin A (§ 33). 
2K=bcsinA. (68) 

2. K= ca sin B, (69) 

and .2 K=ab sin C. (70) 



But in Fig. 1, 
And in Fig. 2, 

Then in either case, 
In like manner, 



Case II. Given one side and all the angles. 
I. When the given parts are a, A, B, and C 
By (70), 2 K=ab sin C. 

b = 

a sin A 
a sin J3 



But by (47), 
Whence, 2 K 
In like manner, 



sin B , a sin JB 
, or b 



a x 



X sin C = 



sin ^4. 

cr sin B sin C 



sin A sin ^4 

6 2 sin C sin J. 



and 



2/f = 

2K= 



sin 5 

c 2 s in ^4 sin B 
sin 



(71) 
(72) 
(73) 



Case III. Given the three sides. 

By (68), 2 if = 6c sin ^4 = 2 6c sin \ A cos J A, by (25). 

Dividing by 2, and substituting the values of sin \ A and cos \ A 
from (59) and (62), we have, 



K. 



b l (s-o)(s-c } l s(s-a) = v,(,_a)(*-&)(*-c). (74) 
^ be V 6c 



SOLUTION OF OBLIQUE TRIANGLES. 67 



VIII. SOLUTION OF OBLIQUE TRIANGLES. 

113. In the solution of oblique triangles, we may distinguish four 
cases. 

114. Case I. Given a side and any two angles. 

The third angle may be found by Geometry, and then by aid of § 107 
the remaining sides may be calculated. 

The triangle is always possible for any values of the given elements, 
provided the sum of the given angles is less than 180°. 

1. Given b = 20.24, A = 103° 36', B = 19° 21' ; find C, a, and c. 

We have, C = 180° -(A + B)= 180° - 122° 57' = 57° 3'. 

-,> /.„\ a sin^l , c sin C 

By (47), -= , and - — 

J v ; ' b smB' b sin B 

Then, a = b sin A esc B, and c = b sin C esc B. 

Whence, log a = log b -+- log sin A -f- log esc B, 

and log c = log b + log sin C + log esc 5. 

log b = 1.306211 log 6 = 1.306211 

log sin A = 9.987649 - 10 log sin C = 9.923837 - 10 

log esc B = 0.479729 log esc B = 0.479729 



log a = 1.773589 log c = 1.709777 

a = 59.3730. c = 51.2598. 

Note) To find log sin 103° 36', take either log cos 13° 36', or log sin 76° 24'. To 
find the log cosecant of an angle, subtract the log sine from 10 — 10. (See Introduc- 
tion to Tables, page viii.) 

EXAMPLES. 

Solve the following triangles : 

2. Given a = 180, ^4 = 38°, 

3. Given b = .82, B = 51° 42' 37", 

4. Given c == 24.637, A = 83° 39', 

5. Given b = .06708, A = 26° 10' 45", 

6. Given a = 5.0454, B = 98° 8' 26", 

7. Given c = 4592.36, A = 74° 27', 

8. Given c = .93109, A = 15° 34' 9", 

9. Given b = 3.67683, A = 67° 21' 54". 
10. Given a = 71396.72, B = 42° 55' 13", C= 16° 4' 57". 



B = 


:75° 


43'. 




(7 = 


= 109 


>°17 


■I oou 


£ = 


:38° 


56'. 




C = 


:44° 


35' 


12". 


c = 


:21° 


51' 


34". 


c = 


= 61° 






= 


123 


°29 


'46". 


5 = 


:57° 


48' 


8". 



68 PLANE TRIGONOMETRY. 

115. Case II. Given two sides and their included angle. 

Since one angle is known, the sum of the other two angles may be 
found ; and then their difference may be calculated by aid of § 108. 

Knowing the sum and difference of the angles, the angles themselves 
may be found; and then the remaining side may be computed as in 
Case I. 

The triangle is possible for any values of the data. 



1. 


Given a = 


82, 


c = 167, B 


= 98° 


14'; 


find A, 0, and b. 


By 


Geometry, 




C+A = 


,180° 


- B 


= 81° 46'. 


T)V 


(52), 




c + a 


tan J 


(C + A) 




c — a 


tan-j 


(0- 


-A) 


Or, 




tan 


HC-A) = 


c — a , 
tan 


\(C + A). 



c + a 

Then, log tan \ (C — A) = log (c — a) + colog (c + a) + log tan i (C + A). 

c - a = 85 log = 1.929419 

c + a = 249 colog = 7.603801 - 10 

i(C + A)= 40° 53' log tan = 9.937377 - 10 

log tan i(0 - A)= 9.470597 - 10 

i(0-^l)=16 o 27'49.8". 

Therefore, C = \(C + A) + \{G - A)= 57° 20' 49.8", 

and A = ±(C + A)- ±(C-A)= 24° 25' 10.2". 

To find the remaining side, we have by (47), 

, a sin B • „ A 

o = — = a sin B esc A. 

smA 

Whence, log b = log a + log sin B + log esc A. 

log a = 1.913814 

log sin B = 9.995501 - 10 

v log esc ^1 = 0.383615 



Solve the following triangles : 

2. Given a = 67, 

3. Given a = 886, 

4. Given b = 4.102, 



log 6 = 


2.292930 


b = 


196.305. 


EXAMPLES. 


gles: 




c = 33, 


B = 36°. 


b = 747, 


C = 71° 54'. 


c = 4.549 


A =62° 9' 38 



SOLUTION OF OBLIQUE TRIANGLES. 69 



5. 


Given a = .5953, 


b = .9639, 


C = 134°. 


6. 


Given b = 1292.1, 


c = 286.3, 


J. = 27° 13'. 


7. 


Given a = 7.48, 


c = 12.409, 


J3 = 83°26'52". 


8. 


Given a = 93.273, 


6 = 81.512, 


C = 58°. 



9. Given b = .0261579, c = .0608657, A = 115° 42'. 
10. Given a = 35384.82, c = 57946.34, B = 19° 37' 25". 

116. Case III. Given the three sides. 

The angles might be calculated by the formulae of § 110 ; but as these 
are not adapted to logarithmic computation, it is usually more convenient 
to use the formulae of § 111. 

Each of the three angles should be computed trigonometrically, for Ave 
then have a check on the work, since their sum should be 180°. 

If all the angles are to be computed, the tangent formulae are the most 
convenient, since only four different logarithms are required. If but one 
angle is required, the cosine formula will be found to involve the least 
work. 

The triangle is possible for any values of the data, provided no side is 
greater than the sum of the other two. 

If all the angles are required, and the tangent formulae are used, it is 
convenient to modify them as follows ; by (65), 



t a ni>l-%/5E^5EMEl_ 1 J (s-a)(s-b)(s-c) 

-rO j(s — a)(s — b)(s — c) , , 

Denoting -v/- — — by r, we have 

* s 

tan \A = 

s — a 

In like manner, tan ^ B = , and tan 4C = 



, UCtii. 2 -LJ , CIXJ.M. UCUJ.X 2 



1. Given a = 2.51, b = 2.79, c = 2.33 ; find A, B, and C. 
Here, 2s = a + b + c = 7.63. 

Whence, s = 3.815, s-a = 1.305, s-b = 1.025, s - c = 1.485. 
We have, log r = \ [log is — a) + log (s — b) + log (s — c) + colog s]. 
Also, log tan \A = log r — log (s — a), 

log tan i B = log r — log (s — 6), 
and log tan ^ C = log r — log (s — c). 



70 PLANE TRIGONOMETRY. 

log (s - a) = 0.115611 log r = 9.858283 - 10 

log (s - b) = 0.010724 log (s - b) = 0.010724 

log {s-c)= 0.171726 log tan i B = 9.847559 - 10 

colog s = 9.418505-10 -i- 5 =35° 8 '40 9" 

2 )19.716566 — 20 2? = 70° 17' 21.8" 

logr = 9.858283-10 . ftQKOOO ' " 

i / % Aiir^n log r = 9.858283 - 10 

l0g (S - G) = °- 115611 log (. - c) = 0.171726 

log tan M = ™ 72 - 10 log tan ^= 9.686557 -10 

\A -28° 56' 22.7". i C= 25°54'56.2". 

^i_o^ oz 4o.4 . 0=51° 49' 52.4". 

Check, A + B + 0= 179° 59' 59.6". 
2. Given a = 7, & = 11, c = 9.6 ; find B. 

By (63), cosi B=yj^^. 

Whence, log cos^B = -J [log s + log (s — 6) + colog c + colog a]. 
Here, 2s = a + b + c = 27.6'; whence, s = 13.8, and s — b — 2.8. 
logs= 1.139879 
log (*-&) = 0.447158 

colog c= 9.017729-10 

colog a = 9.154902-10 

2 ]l9.759668 - 20 

logcosiJ3= 9.879834-10 

15 = 40° 41' 11.5", and B = 81° 22' 23.0". 

EXAMPLES. 

Solve the following triangles : 



3. 


Given a = 2, 


b = 3, 


c = 4. 


4. 


Given a = h, 


b = 7, 


c = 6. 


5. 


Given a = 10, 


b = 9, 


c = 8. 


6. 


Given a = 5.6, 


b = 4.3, 


c = 4.9. 


7. 


Given a = .85, 


b = .92, 


c = .78. 


8. 


Given a = 61.3, 


6 = 84.7, 


c = 47.6. 


9. 


Given a = 705, 


b = 562, 


c = 639 ; find ^. 


10. 


Given a = .0291, 


b = .0184, 


c = .0358 ; find. B. 


11. 


Given a = 3019, 


6 = 6731, 


c-4228; find C. 



SOLUTION OF OBLIQUE TRIANGLES. 71 

117. Case IV. Given two sides and the angle opposite to one of them. 

It was stated in § 97 that a triangle is, in general, completely deter- 
mined when three of its elements are known, provided one of them is 
a side. The only exceptions occur in Case IV. 

To illustrate, let us consider the following example : 

Given a = 52.1, b = 61.2, A = 31° 26'; find B, C, and c. 

r> /at,\ siiiB b Tj b sin A 

By (47), = -, or sin B = 

sini a a 

Whence, log sin B = log b + colog a + log sin A. 

log b = 1.786751 
colog a = 8.283162 - 10 
log sin A = 9.717259 - 10 



log sin B = 9.787172 - 10 ' 

B = 37°46'37.9", from the table. 

But in determining the angle corresponding, attention must be paid 
to the fact that an angle and its supplement have the same sine (§ 33). 

Hence, another value of B will be 180°-37°46'37.9", or 142° 13' 22.1 "'; 
and calling these values B Y and B 2 , we have 

B x -37° 46' 37.9", and 5 2 = 142° 13' 22.1". 

Note. The reason for this ambiguity is at once apparent when we attempt to 
construct the triangle from the data. 







c 


yF 




b/ 


/a 


\a 




n/ 




e \y 


A 


b: 




B t D 



We first lay off the angle DAF equal to 31° 26', and on AF take AC = 01.2. 
With C as a centre, and a radius equal to 52.1, describe an arc cutting AD at 13 l 
and B-2> Then either of the triangles AByC or AB- 2 C satisfies the given conditions. 

The two values of B which were obtained are the values of the angles AB^C 
and AB 2 C, respectively; and it is evident geometrically that these angles are supple- 
mentary. 

To complete the solution, denote the angles ACB { and ACB 2 by Ci 
and C 2 , and the sides AB Y and AB 2 by c v and c 2 . 

Then, C 1 =180 o -(^ + B,)= s 180°- 69° 12' 37.9" = 110° 47' 22.1", 

and C 2 = 180° - (A + B a ) = 180° - 173° 39' 22.1" = 6° 20 '37.9". 



72 PLANE TRIGONOMETRY. 

Again, by (49). c -i = S ™°1, and * = 5SH. 

a sin A a sin A 

Whence, c, = a sin C x esc A, and c 2 = a sin (7 2 esc A 

log a = 1.716838 log a = 1.716838 

log sin d = 9.970761 - 10 log sin C 2 = 9.043343 - 10 

log esc A = 0.282741 log esc A = 0.282741 

log Gi = 1.970340 log c 2 = 1.042922 

c v = 93.3985. c 2 = 11.0388. 

118. Whenever an angle of an oblique triangle is determined from its 
sine, both, the acute and obtuse values must be retained as solutions, 
unless one of them can be shown by other considerations to be inadmissi- 
ble ; and hence there may sometimes be two solutions, sometimes one. and 
sometimes none, in an example under Case IV. 

1. Let the data be a, b, and A, and suppose b < a. 

By Geometry, B must be < A ; hence, only the acute value of B can 
be taken ; in this case there is but one solution. 

2. Let the data be a, b, and A, and suppose b> a. 

Since B must be > A, the triangle is impossible unless A is acute. 

Again, since — = — , and b is > a, sin B is > sin A. 

sin A a 

Hence, both the acute and obtuse values of B are > A, and there are 
two solutions, except in the following cases : 

If log sin B = 0, then sin B = 1 (§ 72), and B = 90°, and the triangle 
is a right triangle; if log sin B is positive, then sinl? is >1, and the tri- 
angle is impossible. 

119. The results of § 118 may be stated as follows : 

If, of the given sides, that adjacent to the given angle is the less, there 
is but one solution, corresponding to the acute value of the opposite angle. 

If the side adjacent to the given angle is the greater, there are two 
solutions, unless the log sine of the opposite angle is or positive ; in 
which cases there are one solution (a right triangle), and no solution, 
respectively. 

120. We will illustrate the above points by examples : 
1. Given a = 7.42, b = 3.39, A = 105° 13'; find B. 

Since b is < a, there is but one solution, corresponding to the acute 
value of B. 



SOLUTION OF OBLIQUE TRIANGLES. 73 

I* /--\ -r> b sin A 
By (47), sin B = 

Whence, log sin B = log b + colog a + log sin A. 

log b = 0.530200 
colog a = 9.129596 -10 
log sin ^1 = 9.984500 -10 

log sin B = 9.644296 - 10 
B = 26° 9' 30.5". 

2. Given 6 = 3, c = 2, C = 100° ; find 5. 

Since 6 is > c, and (7 is obtuse, the triangle is impossible, 

3. Given a = 22.7643, c = 50, ^1 = 27° 5'; find C. 

We have, sin<7 = ^^L 

a 

log c = 1.698970 
colog a = 8.642746 -10 
log sin A = 9.658284 -10 

log sin C = 0.000000 
Therefore, sin (7=1, and (7 = 90°. 

Here there is but one solution ; a right triangle. 

4. Given a = .83, b = .715, B = 61° 47'; find A 

We have, sin^ = asm ^ . 

log a = 9.919078 -10 
^ colog b = 0.145694 

log sin B = 9.945058 - 10 

log sin .4 = 0.009830 

Since log sin A is positive, the triangle is impossible. 

EXAMPLES. 
121. Solve the following triangles : 

1. Given a = 5.98, 6 = 3.59, A = 63° 50'. 

2. Given b = 74.1, c = 64.2, C = 27° 18'. 

3. Given b = .2337, c = .0982, B = 108°. 

4. Given a = 4.254, c = 4.536, C = 37°9'. 

5. Given a = .2789, b = .2271, B = 65° 38'. 



PLANE TRIGONOMETRY 



6. 


Given a = 60.935, 


c = 76.097, 


A = 133° 41'. 


7. 


Given b = 74.8067, 


c = 98.7385, 


C = 81° 47'. 


8. 


Given a = 9.51987, 


c = ll, 


A = 59° 56'. 


9. 


Given b = 4.521, 


c = 5.03, 


B = 40° 32' 7". 


10. 


Given a = 186.82, 


b = 394.2, 


£ = 114° 29' 51". 


11. 


Given b = 5143.4, 


c = 4795.56, 


C = 72° 53' 38". 


12. 


Given a = .860619, 


c = .635761, 


.4 = 19° 12' 43". 


13. 


Given a = 139.27, 


b = 195.9716, 


^1 = 45° 17' 20". 


14. 


Given a = .32163, 


c = .27083, 


(7= 52° 24' 16". 



15. Given b = 91139.04, c = 80640.37, 5=126° 5' 34". 

AREA OF AN OBLIQUE TRIANGLE. 
122. 1. Given a = 18.063, A = 96° 30' 15", JB = 35°0'13"; find K. 

By (71), 2 K = a2sm ^ sme = a 2 sin 5 sin C esc A 

J v ; sin J. 

Whence, log (2K)=2 log « + log sin 1? + log sin C + log esc A. 

From the data, C = 180° -(A + B)= 48° 29 ' 32 ' '. 

log a = 1.256790 ; multiply by 2 = 2.513580 

log sin B = 9.758630 - 10 
log sin C = 9.874404 - 10 
log csc .4 = 0.002804 

log(2K)= 2.149418 

2 K= 141.065. 
7f = 70.533. 

EXAMPLES. 
Find the areas of the following triangles : 

2. Given a = 38.09, e = 11.2, B = 67° 55'. 

3. Given a = 5, 6 = 8, c = 6. 

4. Given b = 6.074, ^ = 70° 39', B = 56° 23'. 

5. Given b = 761.86, c = 526.02, ^4 = 124° 6' 13". 

6. Given a = 97, b = 83, c = 71. 

7. Given a = 1.9375, JL = 43°18', B = 29° 47' 36". 



SOLUTION OF OBLIQUE TRIANGLES. 



75 



8. Given b = .439592, 

9. Given a = 39.5, 
10. Given a = .804639, 



A = 62° 40' * 
b = 44.8, 
c = .357173, 



= 54° 32' 25". 
c - 52.3. 
5 =18° 11' 49". 



11. Given c = 95.86157, 73 = 115° 24' 52", (7= 32° 57' 21". 

12. Given a = .02409481, b = .02763834, C = 81° 9' 34". 

13. Given a = 7.825, b = 6.592, c = 9.643. 



MISCELLANEOUS EXAMPLES. 

123. 1. From a point in the same horizontal plane with the base of 
a tower, the angle of elevation of its top is 52° 39', and from a point 
100 ft. further away it is 35° 16'. Find the height of the tower, and its 
distance from each point of observation. 

2. One side of a parallelogram is 56, and the angles between this side 
and the diagonals are 31° 14' and 45° 37'. Find all the sides of the 
parallelogram. 

3. In a field ABCD, the sides AB, BC, CD, and DA are 155, 236, 
252, and 105 rods, respectively, and the diagonal AC is 311 rods. Find 
the area of the field. 

4. The area of a triangle is 1356, and two of its sides are 53 and 69. 
Find the angle between them. 

5. From the top of a bluff, the angles of depression of two posts in 
the plain below, in line with the observer and 1000 ft. apart, are found to 
be 27° 40' and 9° 33', respectively. Find the height of the bluff above the 
plain. 

6. The parallel sides of a trapezoid are 86 and 138, and the angles at 
the extremities of the latter are 53° 49' and 67° 55'. Find the non-parallel 
sides. 

7. Two trains start at the same time from the same point, and move 
along straight railways, which intersect at an angle of 74° 30', at the rates 
of 30 and 45 miles an hour, respectively. How far apart are they at the 
end of 45 minutes ? 

8. Two sides of a triangle are .5623 and .4977, and the difference of 
the angles opposite these sides is 15° 48' 32". Solve the triangle. 

9. Two yachts start at the same time from the same point, and sail, 
one due north at the rate of 10.44 miles an hour, and the other due north- 
east at the rate of 7.71 miles an hour. What is the bearing of the first 
yacht from the second at the end of half an hour ? 



76 PLANE TRIGONOMETRY. 

10. A vessel is sailing due south-west at the rate of 8 miles an hour. 
At 10.30 a.m., a lighthouse is observed to bear 30° west of north, and at 
12.15 p.m., it is observed to bear 15° east of north. Find the distance of 
the lighthouse from each position of the vessel. 

11. Two sides of a parallelogram are 65 and 133, and one of the 
diagonals is 159. Find the angles of the parallelogram, and the other 
diagonal. 

12. To find the distance of an inaccessible object A from a position B, 
I measure a line BC, 208.3 ft. in length. The angles ABC and ACB are 
measured, and found to be 126° 35' and 31° 48', respectively. Find the 
distance AB. 

13. The diagonals of a parallelogram are 81 and 106, and the angle 
between them is 29° 18'. Find the sides and angles of the parallelogram. 

14. A flagpole 40 ft. high stands on the top of a tower. From a posi- 
tion near the base of the tower, the angles of elevation of the top and 
bottom of the pole are 38° 53' and 20° 18', respectively. Find the distance 
and height of the tower. 

15. AD and BC are the parallel sides of a trapezoid ABCD; the sides 
AB and BC are 7.8 and 9.4, respectively, and the angles B and C are 
113 D 47' and 125° 34', respectively. Find AD and CD. 

16. A surveyor observes that his position A is exactly in line with 
two inaccessible objects B and C. He measures a line AD 500 ft. long, 
making the angle BAD = 60°, and at D observes the angles ADB and 
BDC to be 40° and 60°, respectively. Find the distance BC. 

17. A side of a parallelogram is 48, a diagonal is 73, and the angle 
between the diagonals, opposite the given side, is 98° 6'. Find the other 
diagonal and the other side. 

18. To find the distance between two buoys A and B, I measure a 
base-line CD on the shore, 150 ft. long. At the point C the angles ACD 
and BCD are measured, and found to be 95° and 70°, respectively ; and 
at D the angles BDC and ADC are found to be 83° and 30°, respectively. 
Find the distance between the buoys. 

19. The sides AB, BC, and CD, of a quadrilateral ABCD are 38, 55, 
and 42, respectively, and the angles B and C are 132° 56' and 98° 29', 
respectively. Find the side AD, and the angles A and D. 

20. The sides AB, BC, and DA of a held ABCD are 37, 63, and 20 
rods, respectively, and the diagonals AC and BD are 75 and 42 rods, 
respectively. Find the area of the field. 



CUBIC EQUATIONS. 77 



IX. CUBIC EQUATIONS. 

124. We know, by Algebra, that a cubic equation can always be 
transformed into another in which the term containing the square of the 
unknown quantity shall be wanting. 

Thus, if the equation is x s + px 2 + qx -f r = 0, putting x — y—^, we 
have 

„ 9 , P'V P 3 , o %p~y , P 3 , PQ , « 

which is in the required form. 

125. Cardan's Method enables us to solve any cubic equation of the 

a 3 
form x 2, + ax + b = 0, except in the case where a is negative, and — 

b 2 ^ 

numericallv > — 
4 

In this case, it is possible to find the roots by Trigonometry. 

126. Trigonometric Solution of Cubic Equations. 

To solve the equation 

x^ — ax — b — 0, 

7 ■ a «V o 2 

tohere a is positive, and — > — 

F ' 27 4 

Putting- x = 2 m cos A, the equation becomes 

8 ra 3 cos 3 A — 2 am cos A — b = 0, or 4 cos 3 A cos A = 0. 

rw 2 m 3 

But by (36), 4 cos 3 A = cos 3 A + 3 cos A 

Whence, cos 3 A + 3 cos .4 — -^- cos A = 0. 

m 2 2 m 3 

Or, cos 3 A + /s - ^ cos A = -^ (A) 



We may take m so that 3 — — 2 == ; then, 3 m 2 = a, and wi = a/-. (IV 
Then (A) becomes cos 3 A = 






78 PLANE TRIGONOMETRY. 

Substituting in this the value of m from (B), we have 

eos3A = b -yj^. (C) 

b 2 a? b 2 27 

Since, by hypothesis, — < — , we have — x — < 1. 

Taking the square root of each member of the inequality, -\— < 1. 
Hence, the value of 3 A in (C) is possible, since its cosine is < 1. 
Let z be the least positive angle whose cosine is equal to -\/ — 

Then, one value of 3 A is z ; and all its values are given by the 
expression 2 mr ± z (§ 62), where n is or any positive or negative 
integer. 

Whence, cos A = cos \ (2 mr ± z). 

Now let n = 3q + n\ where q is or any positive or negative integer, 
and n' = or ± 1 ; then, 

A (6ff + 2nV±2 

cos A = cos H J — ±L — — cos 

o 



2?ffH ^~ 



2 ?i'tt ± z 
cos 



3 



for by § 21, any multiple of 360° may be added to, or subtracted from. 
an angle, without altering its functions. 
Putting n' = Q, 1, and — 1, we have 

A f z\ 2ir±Z — 2tt±Z Z 2tt±Z 

cos A = cos ± - , cos — - — , or cos = cos - or cos — - — ; 

\ 3/ 3 3 3 3 

for by § 29, the cosine of the angle (— A) is equal to the cosine of A. 
But x = 2m cos A : and hence the three values of x are 



2 \l c °4 2 4™(¥-i)' and 2 \! c< 



, 3 3/ ' "\3™V3 3j' 
where z is given by the equation cos z =-a/— • 

EXAMPLES. 

1. Solve the equation x 3 — 4 a; + 2 = 0. 

/'>" '97 

Here a = 4, b = — 1' ; then, cos z = — V/-—, or cos (r — 2) =\j — 

^ 04 * 64 

By logarithms, log cos (?r — z) = 1 (log 27 — log 64). 

log 27= 1.431364 
log 64= 1.806180 

2 )19.625184 -"20 
log cos («■ - z) = 9.812592 - 10 



33) 



CUBIC EQUATIONS. 79 

Then, ir - z = 49° 29' 40.5", and z = 130° 30' 19.5". 

Whence, | = 43° 30' 6.5", and 2^| = 2^/| = ^- 

Then the three values of a? are 






cos 43° 30' 6.5", 
3 



—cos (120° - 43° 30'6.5") =-Jl?cos 76° 29' 53.5", 

^ 6 cos (120° + 43° 30' 6.5") - J^ 



and \f^'os (120° + 43° 30' 6.5") = ^/~cos (90° + 73° 30' 6.o") 



16 



^ sin 73° 30' 6.5" (§ 30). 



Now, 



log VT = i( lo o 16 ~ log 3) = 1(1.204120 - .477121) = .363500. (1 ) 
* o 

Also, log cos 43° 30' 6.5" = 9.860549 - 10, (2) 

log cos 76° 29' 53.5" = 9.368242 - 10, (3) 

and log sin 73° 30' 6.5" = 9.981741 - 10. (4) 

Adding (2), (3), and (4) in succession to (1), the logarithms of the 
absolute values of x are 

0.224049, 9.731742 - 10, and 0.345241. 

The numbers corresponding to these logarithms are 

1.67513, .53919, and 2.21432. 
Whence, x = 1.67513, .53919, or -2.21432. 

Solve the following equations : 

2. ar ? -4a-l = 0. * 4. or 3 + 6x 2 - x - 1 = 0. 

3. x i -6x + 3 = 0. 5. x*-3x 2 -2x + l=Q. 



SPHERICAL TRIGONOMETRY. 

X. GEOMETRICAL PRINCIPLES. 

127. If a triedral angle be formed with its vertex at the centre of a 
sphere, it intercepts on the surface a spherical triangle. 

The triangle is bounded by three arcs of great circles, called its sides, 
which measure the face angles of the triedral angle. 

The angles of the spherical triangle are the spherical angles formed by 
the adjacent sides ; and, by Geometry, each is equal to the angle between 
two straight lines drawn, one in the plane of each of its sides, and per- 
pendicular to the intersection of these planes at the same point. 

128. The sides of a spherical triangle are usually expressed in degrees. 

129. Spherical Trigonometry treats of the trigonometric relations 
between the sides and angles of a spherical triangle. 

The face and diedral angles of the triedral angle are not altered by 
varying the radius of the sphere; and hence the relations between the 
sides and angles of a spherical triangle are independent of the length of 
the radius. 

130. We shall limit ourselves in the present work to such triangles 
as are considered in Geometry, where each angle is less than two right 
angles, and each side less than the semi-circumference of a great circle ; 
that is, where each element is less than 180°. 

131. The proofs of the following properties of spherical triangles may 
be found in any treatise on Solid Geometry : 

1. The sum of any two sides of a spherical triangle is greater than 
the third side. 

2. In any spherical triangle, the greater side lies opposite the greater 
angle; and, conversely, the greater angle lies opposite the greater side. 

3. The sum of the sides of a spherical triangle is less than 360°. 

M 



GEOMETRICAL PRINCIPLES. 81 

4. The sum of the angles of a spherical triangle is greater than 180°, 
and less than 540°. 

5. If A'B'C is the polar triangle of ABC, that is, if A, B, and C 
are the poles of the sides B'C, C'A', and A'B', respectively, then, con- 
versely, ABC is the polar triangle of A'B'C. 




6. In two polar triangles, each angle of one is measured by the sup- 
plement of the side lying opposite the homologous angle of the other ; 
that is 

a' =180° -A. b' = lS0°-B. c' = lS0°-C. 

A' = lS0°-a. B' = lS0°-b. C = 180° - c. 

132. A spherical triangle is called tri-rectangular when it has three 
right angles ; each side is a quadrant, and each vertex is the pole of the 
opposite side. 

133. I. Let C be the right angle of the right spherical triangle ABC, 
and suppose a < 90° and b < 90°. 

1 B' 

s7\- 




E 

Complete the tri-rectangular triangle A'B'C-, also, since B' is the pole 
of AC, and A' of BC, construct the tri-rectangular triangles AB'D and 
A'BE. 

Then since B lies within the triangle AB'D, AB or c is < 90°. 

Since BC is < B'C, the angle A is < BAD, or < 90°. 

Since AC is < A'C, the angle B is < A'BE, or < 90°. 



82 SPHERICAL TRIGONOMETRY. 

II. Suppose a < 90° and b > 90°. 




Complete the lune ABA 'C. 

Then in the right triangle A'BC, A'C = 180° - 6. 

That is, the sides a and A'C of the triangle A'BC are each < 90° ; and 
by I., A'B and the angles A' and A'BC are each < 90°. 

But, c == 180° - A'B, A = A', and B = 180° - A'BC 

Whence, c is > 90°, A < 90°, and B > 90°. 

Similarly, if a is > 90° and b < 90°, then c is > 90°, A > 90°, and 
B < 90°. 

Ill, Suppose a > 90° and b >90°. 




b 
Complete the lune ACBC 

Then in the right triangle ABC, AC = 180° - b, and BC = 180° - a. 
That is, the sides AC and BC of the triangle ABC are each < 90' ; 
and by L, AB and the angles BAC and ABC are each < 90°. 
But, A = 180° - BAC, and B = 180° - ABC. 
Whence, c is < 90°, A > 90°, and B > 90°. 
Hence, in any right spherical triangle : 

1. If the sides about the right angle are in the same quadrant, the hypote- 
nuse is < 90° ; if they are in different quadrants, the hypotenuse is > 90°. 

2. An angle is in the same quadrant as its opposite side. 

134. In the figure of § 131, we have, by § 131, 1, a' < b' + c'. 
Putting for a', b', and c' the values given in § 131, 6, we have 

180° - A < 180° - B -f 180° - C, or B-{- C-A<1$0°. 

Again, by § 130, B + C + 180° > A ; whence, B + C - A > - 180°. 

Therefore, B + (7- A is between 180° and - 180°. 

Similarly, O + A - B and .4 + B - (7 are between 180° and - 180°. 



RIGHT SPHERICAL TRIANGLES. 



83 



XL RIGHT SPHERICAL TRIANGLES. 

135. Let C be the right angle of the right spherical triangle ABC. 

B 




Let be the centre of the sphere, and draw OA, OB, and 00. 

At any point A' of OA draw A'B' and A'C perpendicular to OA, meet- 
ing OB and 00 at B' and C, and draw B'C. 

Then OA is perpendicular to the plane A'B'C. 

Hence, each of the planes A'B'C and OBC is perpendicular to the 
plane OAC, and their intersection B'C is perpendicular to OAC. 

Therefore, B'C is perpendicular to A'C and OC. 

In the right triangle OA'B', we have 

/iinT)l OA' OC OA' 

cos c = cos A' OB' — = x 

OB' OB' OC 

But in the right triangles OB'C and OCA', 

OC 4' 

— = cosa, and _=cos&. 

cos c = cos a cos b. 
B'C 



Whence, 



Asjain, 



sin.4 = sin 23M'C" = 



And, cos A = cos B'A'C = 

In like manner, sin B — 



B'C 


OB' 


sin a 


A'B' 


A'B' 
OB' 

A'C 


sine 


A'C 1 
A'B' 


OA' 

~ A'B'' 

OA' 


tan b 
tan c 


sin b 







and 



cosI> 



sine 

tana 
tanc 



(75) 
(76) 

(77) 

(78) 

(79) 



S4 SPHERICAL TRIGONOMETRY. 

136. The proofs of § 135 cannot be regarded as general, for in the 
construction of the figure we have assumed a and b, and therefore c and 
A (§ 133), to be less than 90°. 

To prove formulae (75) to (79) universally, we must consider two addi- 
tional cases : 

Case I. When one of the sides a and b is < 90°, and the other > 90°. 

B 




In the right spherical triangle ABC, let a be < 90° and b > 90°. 
Complete the lune ABA'C\ then, in the spherical triangle A'BC, 

A'B = lS0°-c, A'C=lS0°-b, A' = A, and A'BC=1$0°-B. 

But by § 133, c is > 90°, A < 90°, and B > 90°. 

Hence, each element, except the right angle, of the right spherical 
triangle A'BC is < 90° ; and we have by § 135, 

cos A'B = cos a cos A'C, 



sm A'B smA'B 

A , tan A'C Amn tana 

cos A ' = , cos A'BC 



tan A'B 

Putting for A'B, A'C, A' and A'BC their values, we have 

cos (180° - c) = cos a cos (180° - b), 

a sin a . /i0 Ao t>\ sin (180° — b) 

sin A = — — , sin (180 — B)= . \ — — S 

sin(180°-c)' V ; sin(180°-c)' 

4 tan (180° — b) ,-, OA o -m tan a 

cos A = \- — '-, cos (180° — B) = ——? -• 

tan(180°-c) v ; tan(180°-c) 

Whence, by § 33, — cose = cosa(— cos b), 

A sin a -o sin b 

sin A — , sin B = —. — , 

sin c sm c 

A — tan b -d tan a 

cos A = , — cos B 



tan c — tan c 



and we obtain formulae (75) to (79) as before. 

In like manner, the formulae may be proved to hold when a is > 90 c 
and b < 90°. 



RIGHT SPHERICAL TRIANGLES. 85 

Cask II. When both a and b are > 90°. 

B 

\ c ">c 




ll 



In the right spherical triangle ABC, let a and b be > 90°. 
Complete the lime ACBC 
By § 133, c is < 90°, A > 90°, and B > 90°. 

Hence, each element, except the right angle, of the right spherical 
triangle ABC is < 90°; and we have by § 135, 

cos c = cos AC cos BC, 

sin BAG' = sinBC \ sin ABC = S ™ AC ' , 

sin c sin c 

cos BAC = , cos ABC = 

tan c tan c 

Putting for AC, BC, BAC, and ABC their values, we have 
cos c = cos (180° - a) cos (180° - b), 

sin(l80°-A)=* uU1S0 °- a \ sin(18Q o -^)= sin ^ 180 °-^ 

sin c sm c 

co8(180°-^) = tan(180 °- 6) , cos(180°-.B)= tan(180 °-")- 
tan c tan c 

Whence, by § 33, cos c = (— cos a) (— cos &), 

, sin a . -d sin 6 

sm ./l = — , sm B = , 

sm c sin c 

. — tan b -d — tan a 

— cos^l = , — cosI*= ; 

tan c tan c 

and we obtain formulae (75) to (79), as before. 

137. From (76) and (77), we obtain 

, j _ sin A _ sin a tan c sin a 



cos A sine tan 6 cose tan b 

Whence by (75), tan A = *™J* L tn ^. (&» 

cos a cos b tan 6 sm b 

In like manner, tan B — — (81 1 

sin a 



8Q SPHERICAL TRIGONOMETRY. 

138. By (4), sin a = cos a tan a ; then (76) may be written 

tana 

. cos a tan a tan c 

sm A = - = 

cos c tan c cos c 

cos a 
Whence by (75) and (79), 



A cos 5 

sm A = 

cos b 




In like manner, sin B = C0S • 

cosg 




139. From (75), (82), and (83), we have 




t. cos A cos B 

cos c = cos a cos 6 = — x — = 


cot A cot B 



sinl> sin^l 



(82) 
(83, 

(84) 



140. The formulae of §§ 135 to 139 are collected below for convenience 
of reference: 

cos c = cos g cos b. 

sin ^1 = sin a 

cos A = 

tan A = 

sin A = 

cos b cos a 

cos c = cot A cot i?. 

The student should compare the formulae for the sines, cosines, and 
tangents of A and B with the corresponding formulae in §§ 2 and 5. 

141. Napier's Rules of Circular Parts. 

These are two rules which include all the formulae of § 140. 



co. B 



sin c 


sm c 


tan b 
tan c 


„ tan g 
cos B = 

tan c 


tang 
sin b 


, -d tan 6 

tan 5 = 

sin a 


cos B 


■ -n COS A 

sin jB = 



co. A 




In any right spherical triangle, the elements a and b, and the comple- 
ments of the elements A, B, and c (written in abbreviated form, co. A. 
co. B. and co. c), are called the circular parts. 




RIGHT SPHERICAL TRIANGLES. 87 

If we suppose them arranged in the order in which the letters occur in 
the triangle, any one of the live may be taken and called the middle part ; 
the two immediately adjacent are called the adjacent parts, and the 
remaining two the opposite parts. 

Then Napier's rules are : 

I. The sine of the middle part is equal to the product of the tangents of 
the adjacent parts. 

II. The sine of the middle part is equal to the product of the cosines of , 
the opposite parts. 

142. Napier's rules may be proved by taking each circular part in 
succession as the middle part, and showing that the results agree with the 
formulae of § 140. 

1. If a be taken as the middle part, b and co. B are the adjacent parts, 
and co. c and co. A the opposite parts. 

Then the rules give 

sin a = tan b tan (co. B), and sin a = cos (co. c) cos (co. A). 
Or by § 32, sin a = tan b cot B, and sin a = sin c sin A ; 
which are equivalent to (81) and (76). 

2. If b be taken as the middle part, a and co. A are the adjacent parts, 
and co. c and co. B the opposite parts. 

Then, sin b = tan a tan (co. A) = tan a cot A, 

and sin b = cos (co. c) cos (co. B) = sin c sin B ; 

which are equivalent to (80) and (78). 

3. If co. c be taken as the middle part, co. A and co. B are the adjacent 
parts, and a and b the opposite parts. 

Then, sin (co. c) = tan (co. A) tan (co. B), and sin (co. c) = cos a cos b. 

Or, cos c = cot A cot B, and cos c = cos a cos b ; 

which agree with (84) and (75). 

4. If co. A be taken as the middle part, b and co. c are the adjacent 
parts, and a and co. B the opposite parts. 

Then, sin (co. A) = tan b tan (co. c), and sin (co. A) = cos a cos (co. B). 

Or, cos A = tan b cot c, and cos A = cos a sin B ; 

which are equivalent to (77) and (83). 



88 SPHERICAL TRIGONOMETRY. 

5. If co. B be taken as tlie middle part, a and co. c are the adjacent 
parts, and & and co. A the opposite parts. 

Then, sin (co. B) = tan a tan (co. c), and sin (co. B) = cos b cos (co. A). 

Or, cos B = tan a cot c, and cos B = cos b sin A ; = 

which are equivalent to (79) and (82). -- ■ y({ 

Writers on Trigonometry differ as to the practical value of Napier's 
rules; but in the opinion of the highest authorities, it seems to be re- 
garded as preferable to attempt to remember the formulae by comparing 
them with the analogous formulae for plane right triangles, as stated in 
§140. 

SOLUTION OF RIGHT SPHERICAL TRIANGLES. 

143. To solve a right spherical triangle, two elements must be given 
in addition to the right angle. 

There may be six cases: 

1. Given the hypotenuse and an adjacent angle. 

2. Given an angle and its opposite side. 

3. Given an angle and its adjacent side. 

4. Given the hypotenuse and another side. 

5. Given the two sides a and b. 

6. Given the two angles A and B. 

144. Either of the above cases may be solved by aid of § 140. 

The formula for computing either of the remaining elements when any . 
two are given may be found by the following rule : 

Take that formula which involves the given parts and the required part. 

If all the remaining elements are required, the following rule may be 
found convenient in selecting the formulae : 

Take the three formulae which involve the given parts. 

145. It is convenient in the solution to have a check on the log- 
arithmic work, which may be done in every case without the necessity of 
looking out any new logarithms. 

Examples of this will be found in § 148. 

The check formula for any particular case may be selected from the 
set in § 140 by the following rule : 

Take that formula which involves the three required parts. 

Note. If Napier's rules are used, the following rule will indicate which of the 
circular parts corresponding to the given elements and any required element is to be 
regarded as the middle part. 



RIGHT SPHERICAL TRIANGLES. 89 

//' these three circular parts are adjacent, take the middle one as the middle part, 
and the others are then adjacent parts. 

If they are not adjacent, take the part ichich is not adjacent to either of the others 
as the middle part, and the others are then opposite parts. 

For the check formula, proceed as above with the circular parts corresponding to 
the three required elements. 

Thus, if c and A are the given elements, 

1. To find a, consider the circular parts a, co. c, and co. A ; of these, a is the 
middle part, and co. c and co. A are opposite parts. Then, by Napier's rules, 

sin a = cos (co. c) cos (co. A) = sin c sin A. 

2. To find b, the circular parts are b, co. c, and co. A ; in this case co. A is the 
middle part, and b and co. c are adjacent parts. Then, 

sin (co. ^4) = tan b tan (co. c), or cos A = tan b cot c. 

8. To find B, the circular parts are co. B, co. c, and co. A ; co. c is the middle part, 
and co. A and co. B are adjacent parts. Then, 

sin (co. c) = tan (co. A) tan (co. J5), or cos c = cot A cot B. 

4. For the check formula, the circular parts are a, b, and co. B ; a is the middle 
part, and b and co. B are adjacent parts. Then, 

sin a = tan b tan (co. B) = tan b cot B. 

146. In solving spherical triangles, careful attention must be given to 
the algebraic signs of the functions ; the cosines, tangents, and cotangents 
of angles between 90° and 180° being taken negative (§ 20). 

It is convenient to place the sign of each function just above or below 
it, as shown in the examples of § 148 ; the sign of the function in the first 
member being then determined in accordance with the principle that like 
signs produce -f, and unlike signs produce — . 

Note. In the examples after the first of § 148, the signs are omitted in every case 
where both factors of the second member are -f . 

147. In finding the angles corresponding, if the function is a cosine, 
tangent, or cotangent, its sign determines whether the angle is acute or 
obtuse ; that is, if it is +, the angle is acute ; and if it is — , the angle is 
obtuse, and the supplement of the acute angle obtained from the tables 
must be taken (§ 33). 

If the function is a sine, since the sine of an angle is equal to the sine 
of its supplement (§ 33), both the acute angle obtained from the tables 
and its supplement must be retained as solutions, unless the ambiguity 
can be removed by the principles of § 133. 



90 



SPHERICAL TRIGONOMETRY. 



/; 



EXAMPLES. 

148. 1. Given B = 33° 50', a = 108° ; find A, b, and c. 
By the rule of § 144, the formulae from § 140 are, 

. -o cos A , -r, tan b -, ^ tana 

sm B = , tan B = — , and cos B 



cos a 



sm a 



tanc 



~ ~~ + + + ' — £ an a 

Or, cos A = cos a sin B, tan b = sin a tan B, and tan c = 

cos 5 

Hence, log cos A = log cos a + log sin B. 

log tan & — log sin a + log tan B. 
log tan c = log tan a — log cos B. 

Since cos A and tanc are negative, the supplements of the acute angles 
obtained from the tables must be taken (§ 147). 

Note 1. When the supplement of the angle obtained from the tables is to he 
taken, it is convenient to write 180° minus the element in the first member, as shown 
below in the cases of A and c. 

By the rule of § 145, the check formula for this case is 

cos A = , or log cos A = log tan b — log tan c. 

tanc 

The values of log tan b and log tanc may be taken from the first part 
of the work, and their difference should be equal to the result previously 
found for log cos A. 



log cos a 
log sin B 



9.489982 - 10 
9.745683 - 10 



log tan a = 0.488224 

log cos B = 9.919424 - 10 



log 


cos A = 


-. 9.235665 - 


10 


180 ( 


>-A = 


: 80° 5' 33.8" 






A = 


: 99° 54' 26.2 


it 


log 


sin a = 


: 9.978206 - 


10 


log 


tan B- 


: 9.826259 - 


10 



log tan b 



9.804465 - 10 
32° 30' 59.8". 



log tanc =0.568800 
180° -c= 74° 53' 45.0". 
c = 105° 6' 15.0". 

Check. 
log tan b = 9.804465 - 10 
log tanc =0.568S00 

log cos^l = 9.235665 -10 



2. Given c = 70° 30', A = 100° ; find a, b, and B. 
In this case the three formulae are 



sin A 



sin a 

sin c 



cos A 



tan b 
tan c 



and cos c = cot A cot B. 



- + - + 

Or, sin a = sin c sin A, tan b = tan c cos A, and cot B = cos c tan A 



RIGHT SPHERICAL TRIANGLES. 91 

Here the side a is determined from its sine; but the ambiguity is 
removed by the principles of § 133 ; for a and A must be in the same 
quadrant. Therefore, a is obtuse; and the supplement of the angle 
obtained from the tables must be taken. 

By § 145, the check formula is 

tan B = — — , or sin a — tan b cot B. 

sm« 

Note 2. The check formula should always be expressed in terms of the functions 
used in determining the required parts ; thus, in the case above, the check formula is 
transformed so as to involve cot B instead of tan B. 



log sine =9.974347-10 
log sin .4 = 9.993351 -10 


log cose =9.523495-10 
log tan ^4 = 0.753681 ' 


log sin a =9.967698-10 


log cot B = 0.277176 


180° -a = 68° 10 '28.2". 
a = 111° 49' 31.8". 


180° -£ = 27° 50 '39.8". 
£=152° 9 '20.2". 


log tan c = 0.450851 

log cos A = 9.239670 - 10 


Check. 


log tan b = 9.690521 - 10 
180° -b = 26° V 18.4". 


log tan b =9.690521-10 
log cot 5 = 0.277176 



b = 153° 52' 41.6". log sin a = 9.967697 - 10 

Note 3. We observe here a difference of .000001 in the two values of log sin a. 
This does not necessarily indicate an error in the work, for such a small difference 
might easily be due to the fact that the logarithms are only approximately correct to 
the sixth decimal place. 

3. Given a = 132° 6', b = 77° 51' ; find A, B, and c. 
In this case the three formulae are 

. ~ A tana . „ tan b -, ~ ~ "*", 

tan A = , tan B = — — , and cos c = cos a cos b. 

sin b sin a 

+ 
The check formula is 

cos c = cot A cot B, or cos c tan A tan B = l, 

That is, log cos c 4- log tan A + log tan B = log 1=0. 

log tan a = 0.044039 log cos a = 9.826351 - 10 

log sin b = 9.990161 - 10 log cos b = 9.323194 - 10 

log tan A = 0.053878 log cos c = 9.149545 - 10 

180° - A = 48° 32' 41.8". 180° - c = 81° 53' 17.4". 

A = 131° 27' 18.2". c = 98° 6' 42.6". 



92 



SPHERICAL TRIGONOMETRY. 



log tan 6 

log sin a 

log tan B 
B 



0.6669G7 
9.870390 



10 



0.796577 
80° 55' 26.6". 



Check. 
log cose =9.149545-10 
log tan .4 = 0.053878 ■ 
log tan B = 0.796577 

log 1 = 0.000000 



4. Given .1 = 105° 59', a 
The formulae are 



128° 33'; find b, B, and c. 



+ 

sin b 



tana 
tan A' 



+ 
sin B 



cos^L 
cos a 



, and sine 



sin a 
sin A 



The check formula is sinjB: 



sin b 
sin c 



In this case, each of the required parts is determined from its sine ; 
and as the ambiguity cannot be removed by § 133, both the acute angle 
obtained from the tables and its supplement must be retained in each 
case. 

log tan a =0.098617 log sin a =9.893243-10 

log tan A = 0.542981 log sin A = 9.982878 - 10 



log sin b =9.555636-10 


log sine =9.910365-10 


b = 21° 3' 58.7" 


c = 54° 26' 26.7" 


or 158° 56' 1.3". 


or 125° 33' 33.3". 




Check. 


log cos A = 9.439897 - 10 


log sin b =9.555636-10 


log cos a =9.794626-10 


log sine =9.910365-10 


log sin B =9.645271-10 


log sin B =9.645271-10 



B = 26° 13' 18.2" 
or 153° 46' 41.8". 

It does not follow, however,, that these values can be combined pro- 
miscuously ; for by § 133, since a is > 90°, with the value of b less than 
90° must be taken the value of c greater than 90°, and the value of B less 
than 90° ; while with the value of b greater than 90° must be taken the 
value of c less than 90°, and the value of B greater than 90°. 

Thus the only solutions of the example are : 

1. b = 21° 3' 58.7", c = 125° 33' 33.3", £ = 26° 13' 18.2". 

2. 6 = 158° 56' 1.3", c = 54° 26' 26.7", B = 153° 46' 41.8". 



RIGHT SPHERICAL TRIANGLES. 93 

Note 4. The ligure shows geometrically why there are two solutions in this case. 

B 



">a! 



For if AB and AC be produced to A', forming the lune ABA'C, the triangle 
A'BC has the side a and the angle A' equal, respectively, to the side a and the angle 
A of the triangle ABC, and both triangles are right-angled at C. 

It is evident that the sides A'B and A'C and the angle A'BC are tke supplements 
of the sides c and b and the angle ABC, respectively. 

Solve the following right spherical triangles : 

5. Given c = 49°, a = 27°. 

6. Given A = 38°, B = 63°. 

7. Given A = 31°, a = 23°. 

8. Given B = 153°, a = 35°. 

9. Given a = 15°, 6 = 106°. 

10. Given c = 139°, ^ = 165°. 

11. Given B = 82° 25', 6 = 68° 35'. 

12. Given c = 75°37', JB = 29°4'. 

13. Given c = 118°49', 6 = 44° 23'. 

14. Given a = 171° 6', b = 161° 58'. 

15. Given i? = 100° 40', a = 170° 38'. 

16. Given A = 102° 57', B = 143° 46'. 

17. Given a = 10° 28', b = 7° 10'. 

18. Given A = 54° 11', 6 = 83° 29'. 

19. Given ^4 = 50° 43', B = 122° 18'. 

20. Given c = 59°3', A = 147° 32'. 

21. Given 5 = 103° 30', b = 132° 54'. 

22. Given ^4 = 95° 15', b = 166° 7'. 

23. Given c = 78°52', a - 114° 26'. 

24. Given c = 127°9 r , J3 = 80°51'. 

25. Given A = 98° 34', a = 113° 12'. 

26. Given c = 136°21', b = 157° 41'. 



Q4/ SPHERICAL TRIGONOMETRY. 

///6-03 -Itl 

I 149. Quadrantal Triangles. 

A spherical triangle is called quadrantal when it has one side equal to 
a quadrant. 

By § 131, 6, the polar triangle of a quadrantal triangle is a right 
spherical triangle. 

Hence, to solve a quadrantal triangle, we have only to solve its polar 
triangle, and take the supplements of the results. 

1. Given c = 90°, a = 67° 38', b = 48° 50' ; find A, B, and C. 
Denoting the polar triangle by A'B'C, we have by § 131, 6 : 

C" = 90°, A' = 112° 22', B' = 131° 10' ; to find a', 6', and c\ 
By § 144, the formulas for the solution are 

, cos .4' ~~ ,. cos B' -, + , ~ .. ,~, 

cosa' = , coso' = - -, and cos c = cot A' cot B'. 

sin B' sm A' 

+ + 

The check formula is cos c' = cos a' cos 6'. 

log cos A' = 9.580392 - 10 log cot A' = 9.614359 - 10 

log sin B' = 9.876678 - 10 log cot B' = 9.941713 - 10 



log cos a' =9.703714-10 log cose' =9.556072-10 

180° - a' = 59° 38' 9.7". c' = 68° 54' 41.5". 

log cos B' = 9.818392 - 10 Check. 

log sin A' = 9.966033 - 10 log cos a' = 9.703714 - 10 

log cos b< = 9.852359 -"lO lo S cos b ' = 9-852359-10 

180° - b' = 44° 37' 5.8". log cos c' = 9.556073 - 10 

Then in the given quadrantal triangle, we have 

.4 = 180° -a' = 59° 38' 9.7", 

B = 180° -b' = 44° 37' 5.8", 

C = 180° -c'= 111° 5' 18.5". 

EXAMPLES. 
Solve the following quadrantal triangles : 

2. Given A = 122°, b = 154°. 

3. Given A= 45° 52', B = 139° 24'. 

4. Given a= 30° 19', C= 42° 31'. 

5. Given B = 51° 35', C = 116° 13'. 

6. Given A = 105° 8', a = 104° 56'. 

7. Given a= 67° 27', 6= 81° 40'. 



RIGHT SPHERICAL TRIANGLES. 95 

150. Isosceles Spherical Triangles. 

We know, by Geometry, that if an arc of a great circle be drawn from 
the vertex of an isosceles spherical triangle to the middle point of the 
base, it is perpendicular to the base, bisects the vertical angle, and divides 
the triangle into two symmetrical right spherical triangles. 

By solving one of these, we can find the required parts of the given 
triangle. 

1. Given a = 115°, b = 115°, C = 71° 48' ; find A, B, and e. 

Denoting the elements of one of the right triangles by A', B', C", a', 
b' } and c', where C is the right angle, we have 

c' = a = 115°, and A' = ± C = 35° 54'. 

We have then to find the parts a' and B' in this triangle. 

By § 140, sin A' = ^^, and cos c' = cot A' cot B'. 
sin c 

- + 



log cos c' = 9.625048 - 10 
log ten. A' = 9.859666 - 10 

log cot B' = 9.485614 - 10 
180° - B' == 72° 59' 23.5". 
B' = 107° 0' 36.5". 
Then in the given isosceles triangle, 

A = B=B' = 107° 0' 36.5", and c = 2 a' = 64° 12' 17.2". 

EXAMPLES. 

Solve the following isosceles spherical triangles : 

2. Given A = 27° 12', B = 27° 12', c = 135° 20'. 

3. Given a = 152° 6', b = 152° 6', C= 67° 46'. 

4. Given a -112° 25', & = 112°25' ? c = 123°48'. 

5. Given A = 159°, B = 159°, a = 137° 39'. 






% 




sin a' — sin c' sin A 




log 


sin 


c' 


= 9.957276 - 


-10 




log 
log 


sin 
sin 


A' 
a' 


= 9.768173 - 


-10 




= 9.725449 - 


-10 








a' 


= 32° 6' 8.6". 





96 



SPHERICAL TRIGONOMETRY. 






XII. OBLIQUE SPHERICAL TRIANGLES. 



, / tt M • « GENERAL PROPERTIES OF SPHERICAL TRIANGLES. 

151. In any spherical triangle, the sines of the sides are proportional to 
the sines of their opp'osite angles. 





Let ABC be any spherical triangle, and draw the arc CD perpen- 
dicular to AB. 

There will be two cases according as CD falls upon AB (Fig. 1), or 
upon AB produced (Fig. 2). 

In the right triangle ACD, in either figure, we have 

sin CD 







sin A = . , by (76). 
sin b 


Also, in Fig. 1, 




■ r> sin CD 

sin B = — — 

sin a 


And in Fig. 2, 




sin B = sin (180° - CBD) 
= sin CBD (§ 33) = S 




Dividing these 


equat 


ions, we have in either case 

sin CD 
sin A sin b sin a 



sin CD 
sin a 



In like manner, 



and 



sin B sin CD sin b 

sin a 

sin i? _ sin b 

sin sin c 

sin ^4 _ sin a 
sin C sin c 



(85) 

(86) 
(87) 



l^ 



152. In any spherical triangle, the cosine of any side is equal to the 
product of the cosines of the other two sides, plus the continued product of 
their sines and the cosine of their included angle. 



OBLIQUE SPHERICAL TRIANGLES. 97 

In the right triangle BCD, in Fig. 1, § 151, we have, by (75), 
cos a = cos BD cos CD = cos (c — AD) cos CD. 

And in Fig. 2, 

cos a = cos BD cos CD = cos (AD — c) cos CD. 

Whence, in either case, by (12), 

cos a = cos c cos AD cos CD -f sin c sin AD cos CD. 

But in the right triangle A CD, 

cos AD cos (7Z> = cos b, by (75). 

And, sin AD cos CD = sin AD = cos b tan AD 

cos AD 

, tan AD • 7 ^ , /t ,„ x 

= sin 6 = sin 6 cos A, by (77). 

tan 6 ' J K J 

Whence, cos a = cos b cos c + sin 6 sin c cos A. (88) 

In like manner, cos b = cos c cos a + sin c sin a cos B, (89) 

and cos c = cos a cos 6 + sin a sin & cos C. (SO) 

153. Let ABC and A'B'C be a pair of polar triangles. 



X 




Applying formula (88) to the triangle A'B'C, we obtain 

cos a' = cos b' cos c' -f- sin 6' sin c' cos A'. 
Putting for a', b', c', and A' the values given in § 131, 6, we have 
cos (180° -A) = cos (180° - B) cos (180° - C) 

4- sin (180° - B) sin (180° - C) cos (180° - a). 

Whence, — cos A = (— cos. 2?) (— cos (7) 4- sin B sin (7(— cos a) (§ 33). 

That is, cos A — — cos 5 cos (7 + sin 5 sin C cos a. (91) 

Similarly, cos B = — cos (7 cos A + sin C sin A cos b, (92) 

and cos (7 = — cos A cos i? -+- sin A sin 5 cos c. (93) 

The above proof illustrates a very important application of the theory 
of polar triangles to Spherical Trigonometry. 



98 SPHERICAL TRIGONOMETRY. 

If any relation has been found between the elements of a spherical 
triangle, an analogous relation may be derived from it, in which each side 
or angle is replaced by the opposite angle or side, with suitable modifica- 
tions in the algebraic signs. 

154. To express the sines, cosines, and tangents of the half-angles of a 
spherical triangle in terms of the sides of the triangle. 

From (88), § 152 sin b sin c cos A = cos a — cos b cos c. 

,, r , ., cos a — cos b cos c , * x 

W hence, cos A = — - — : (A) 

sin b sin c 

Subtracting both members from 1, we have 
H A i cos a — cos b cos c cos b cos c + sin b sin c — cos a 

1 — COS A = I ; ; = ; 

sin b sin c sin b sin c 

i , . . 01 A cos (b — c) — cos a 

Whence, by (31), 2 sin 2 i A = . , ; . 

' J v ' z sin b sin c 

But by (20), cos y — cos x = 2 sin \ (x + y) sin -J- (x — y). (B) 

__, o ■ 2 1 x 2 sin i [a + (b — c)] sin i [a — (6 - c)] 

Whence, 2 snr A ^L = — — . v 1 . — — /J , 

^ sin 6 sin c 

. „ , . sin 1 (a + & — c) sin i (a — b + c) 
or, sin 2 \A = l — = — {—. — ^ '-> 

1 sin b sin c 

Denoting the sum of the sides, a + b + c, by 2 s, we have 
a + & _ c = ( a + 5 + c )_ 2 c = 2 s — 2 e = 2 (s — c), 
and a-&-f-c=(a + & + c)-2& = 2s-2& = 2(s-&). 

. „ , . sin (s — b) sin (s — c) 

Whence, sm 2 hA = v . ^ . L • 

^ sm b sm c 

^ • , a ^ /sin (,s — b) sin (s — c) , _. 

Or, smli=V v . I . v ^ (94) 

' ^ * sin 5 sm c ' 

t r, i t> ^ /sin (s — c) sin (s — a) / Q _ N 

In like manner, sin i JB = V — • — ^— ^ ~* t 95 ) 



and sin i 



c ^ / sin (s - a) sin (s - fr) 
V sin a sin b 



Again, adding both members of (A) to 1, we have 

cos a — cos b cos c cos a — (cos b cos c — sin b sin c) 

l+C0SA=l-\ : =— -. = : j : - 

sm b sm c sm b sm c 

™n_ u /o«n o " 1 a cos a — cos (& + c) 

Whence, by (32), 2 cos- \A= = — =— ^ '- 

' J v n l sm o sm c 



OBLIQUE SPHERICAL TRIANGLES. 99 

rrru u /T>\ O 2 1/1 2 sin 1 (& + C + Cl) sill i (6 + C — tt) 

Then by (B), 2 cos- -^ J. = ^ : — ^—. — ^— '-. 

J v y L sin o sm c 

Putting a + b + c = 2s, whence b + c — a = 2 (s — a), we have 

„ , . sin s sin (s — a) 

cos- A A = = — =-A ^. 

sm o sm c 



» . < /sin s sin (s — a) 

Or, cos J xl = \J — o .^ h \ n n J > (97 ) 



sin b sin c 



t vi i t> ^ /sm s sin (s — 6) 

In like manner, cos A B =\ : ^ S 

z * sm c sm a 



. ~ /sm s sin (s — c) 

and cosi<7=\ = ^ — ; — -• 

' * sm a sm 6 

Dividing (94) by (97), we have 

tan i A J sin <> - 6) sin (s - c) J sin b sine 

* sin 6 sin c * sin s sin (s — a ) 



(98) 
(99) 



Vsin (s — b) sin (s — c) 
r • f ^ , } ' (100) 



sin s sin (s — a) 



t n i t> sin (s — c) sm (,s — a) /,«-,\ 

In like manner, tan 4- 5= A t / , JN — -, (101) 

^ * sin .<? sin f .<? — /)» 



sin s sin (s — b) 



. ~ /sin (s — a) sin (s — 6) , 

and tan ± C =\ t / , v , • (102) 

L * sm s sm (s — c) / 



*s 155. ^ To express the sines, cosines, and tangents of the half-sides of a 
spherical triangle in terms of the angles of the triangle. 

From (91), § 153, sin B sin C cos a = cos A + cos B cos 0. 

Whence, cos a = cos A + cos B cos °. (A) 

r„, -, i cos xl + cos B cos C 

Then, 1 — cos a — 1 — : — 

sm B sm C 

r. o • 21 — (cos 2? cos C — sin .B sin C) — cos -4 

Or, 2 silica = —^ : — . £ 

sm 5 sin C T 

_ cos (B + C) + cos xl 
sin J3 sin C 

m u /1Q x n • m 2cosi(^ + C + ^)cosi(5+(7-^) 

Then by (19), 2 sm- i a = *-± = — ^-= — ^ '-* 

J K n 2 smBsmC 



Iff SPHERICAL TRIGONOMETRY. 

1 )enoting the sum of the angles, A + B + C, by 2 S, we have 
5+ (7- .4 = 2(^-^4). 

Whence, sin^a = - C0S ? cos ^ ~ ^ • 

sin S sin C 



Or, B m|a=J-^ cos ^-^). (103; 

\ sin R sin f7 



In like mannner, smjb =J- cosS cos( - S ~ B \ (104, 

* sin *7 sin A 



and sin i c= J_gg^cos(S-OX ( 105) 

* sin ^1 sin B 

Again, adding both members of (A) to 1, we have 

1 _-, , cos^l + cos B cos C _ cos^l + cos B cos (7+ sin B sin C 

J. -J— COS ft J. -j ; ; — ; ; — ■ 

sin B sin 6 sin B sin C 

Then, 2 eos'Ia = cos ^ + cos (ij - C) 

sin B sin (7 

_ 2 cosi[JL + £ - C] cos^ -(£ - C)X 
sin I? sin C 

Or, cog ,, a = coai(^ + J-C)eosi(^-B+C) . 

sin i? sin C 

VmtA + B-C=2(S-C), and ^ - B + C r = 2(# - B). 

Whence, cosU a = ™s(S - B) cos(S - C\ 

sin jB sin C T 



~ , cos (aS — B) cos (£ — (7) /_ rt -s 

Or, cos i a = y\ * ■ L ■ n ( 106 ' 

* sm B sin (7 



In like manner, cos ^ = J eos(g- C) eos(5 - A) ^ 

* sin G sin -4 



and cos ^ c 



/cos (S-^) cos Off -3) . (108) 

* sin A sin 7? 



Dividing (103) by (108), we have 



tenia =J ™sSeo S (S-A) ( } 

2 \ eos(S -B) cos (S-C) 



In like manner, t^b=yl cos 6' cos (g -B) (llQ 

* cos (S — C) cos (S — A) 



and tanlc=J eosScos(S-C) (ra) 

2 \ r-osftf- /hoosfN- 7r> 



cos^-^cos^-U) 



OBLIQUE SPHERICAL TRIANGLES. 1<H 

&$ NAPIER'S ANALOGIES. 

156. Dividing (100) by (101), we have 
tan 



)/ I s !- 



\A _ /sin (s — b) sin (s — c) / sins sin (s — 6) 
1-7? \ sin s sin (ft — rA * sin f.s — <Vl sin (s — 



tan ±B * sin s sin (s — a) * sin (s — c) sin (s — a) 



\ B 



sin -J- ^4 cos ^ B /sin 2 (s — b) sin (s — b) 



Or 

cos -J- ^4 sin \B \ sin 2 (s — a) sin (s — a) 

Whence by composition and division, 

sin \ A cos \ B -f- cos-^-^1 sin-i-i?_ sin(s — 6)+ sin(s — a) 
sin -i-^l cos -J- -B — cos|-^4 sin^i? sin (s — 6) — sin (s — a) 

Then by (9), (11), and (21), 

sin (\A + \B)_ tani[s — b -f s — a] 
sin (J^A — -J- jB) ~~ tan -J- [s — b — (s — a)] 

But s — b + s — a = 2s — a — b = c. 

, xr , sin I (A + B) tan J- c , 

Whence, -^ ? ) . D ; = rT -? — =-• (112) 

sm \{A — B) tan i (a — b) x 

157. Multiplying (100) by (101), we have 

tanM tanjB= J 8in ^ ~ 6) f (S ^ °> J sin( > ~ c > ^ F^l. 
L * sm s sm (s — a) * sm s sm (s — b) 



( sin i A sin 1 5 _ /sin 2 (s — c) _ sin (s — c) 

cos^-J.cosi-JB — \ sin 2 s sins 

Whence by composition and division, 

cos J .4 cos -J-5 — sin -J- -4 sin J 2? _ sins — sin(s — c) 
cos-j-,4 cos|-I? -f- sin-J-^1 sin^-i? - sins + sin(s — c) 

0r w tei^ cosG A + \B) _ tan ^[a -(s - c)] 

But 



cos(|^l- JJB) 


tan i [s -47 


s — c] 


s + s — c = 


2s — c = a 


+ b. 


cosi(J. + 5) 


tanic 





Whence, f±- t ^t = t — , , , , N - (113) 

cos i (^L — i?) tan^(a + 6) 

158. Applying formula (112) to the triangle A'B'C, in the figure of 
153, we obtain 

sm%(A' + B 1 ) _ tanic' 
sini(^.'- B')~taiU(a'-&')' 




1#2 SPHERICAL TRIGONOMETRY. 

But, \ (A' + B') = i (180° - a + 180° - b) = 180° - -J (a + &) ; 
%(A' - B') =|(180 o - a - 180° + &) = - |(a - 5) ; 
ic' = i(180°-C)=90°-i-(7; 
and ^( a ' - 6')= i(180° - A - 180° + B)= - %(A - 5). 

Whence ?? [ 180 ^Z ife±*)] = tan (90° - | C) 

sin[-i(a-6)] tan [-K^ -■»)]■ 

Therefore, by §§ 29, 32, and 33, 

sin \ (a + 6) cot -J- (7 

- sin£(a - b) ~ - tan \{A - B) ,|^ 

Or sini(g + 6) _ cotfrC /n^J> 

sin i (a -6) tani(^l-J3)' v ; 

In like manner, from (113), we obtain 

cosj^J-ZT)_ _tanj : ^_ 
cosi(^'-5') _ tani(a' + b')' 

But, i(a' + 6')= 1(180° - -4 + 180° - .B) = 180° -\(A + £). 

W l 1Pn pp cos[180°-i(^ + ^)] _ tan(90°-jC) 

' cos [- i(a - b)~] ~ tan [180° -$(A + B)j 

Therefore, by §§ 29, 32, and 33, 

— cos \ (a -f- b) _ cot i C 

cos -J (a — b) ~ — tan -J- (^4 + B)' 

cos$(a+b) coUC ril _, 

cosi(a-6) tan ^(^1 + 5)' ^ ; 

159. The formulse exemplified in §§ 156, 157, and 158 are known as 
Napier's Analogies. In each case there may be other forms according 
as other elements are used. 

SOLUTION OF OBLIQUE SPHERICAL TRIANGLES. 

160. In the solution of oblique spherical triangles, we may distinguish 
six cases : 

1. Given a side and the adjacent angles. 

2. Given two sides and their included angle. 

3. Given the three sides. 

4. Given the three angles. 

5. Given two sides and the angle opposite to one of them. 
G. Given two angles and the side opposite to one of them. 






OBLIQUE SPHERICAL TRIANGLES. If 3 

By application of the principles of § 131, 6, the solution of an exam- 
ple under Case 2, 4, or 6, may be made to depend upon the solution of an 
example under Case 1, 3, or 5, respectively ; and vice versa. 

Hence, it is not essential to consider more than three cases in the 
solution of oblique spherical triangles. 

The student must carefully bear in mind the remarks made in §§ 146 
and 147. 

161. Case I. Given a side and the adjacent angles. 

1. Given A = 70°, B = 132°, c = 116°; find a, b, and C. 

By Napier's Analogies (§§ 156, 157), we have 

sin \ (B -f A) _ tan \ c , cos \{B + A) _ tan i c 

sin i (B - A) ~ tan £ (b — a)' an cos i(B — A) ~ tan £ (b + a)' 

Whence, tan \ (b — a) = sin 1 (B — A) esc \ {B -f A) tan \ c, 

+ - + 

and tan J (6 + a) = cos 1 (i? — .4) sec -*- (B + -4) tan 1 c. 

From the data, %(B- A) = 31°, J (B + A) = 101°, £ c - 5S°. 

log sin i (JB - A) = 9.711839 - 10 log cos \ (B - A) = 9.933066 - 10 
log esc \(B + A) = 0.008053 log sec \ (B + A) = 0.719401 

log tan \ c = 0.204211 log tan \ c = 0.204211 



log tan 1 (b - a) = 9.924103 - 10 log tan | (6 + a) = 0.856678 

1 (6 - a) = 40° 1' 7.7". 180° - \ (b + a) = 82° 4' 51.8". 

1.(6 + o) = 97° 55' 8.2". 

Then, a = 1 (6 + a) - \ (b - a) = 57° 54' 0.5", 

and 1 6 = 1 (6 + a) + \ (b - a) = 137° 56' 15.9". 

To find (7, we have by § 158, 

cot £ C = s | n |^ a ^ tan J (B-^) = sin | (6 + ct) esc 1 (6 -a) tan 1 (B--4) 

log sin 1 (6 + a) = 9.995839 - 10 
log esc \ (b - a) = 0.191763 
log tan 1 (£ - .4) = 9.778774 - 10 

log cot i C = 9.966376 - 10 

i(7= 47° 12' 56.7", and (7= 94° 25' 53.4". 



Note 1. The value of C may also be found by the formula 

Note 2. The triangle is possible for any values of the given elements. 



104 SPHERICAL TRIGONOMETRY. 

EXAMPLES. 
Solve the following spherical triangles : 

2. Given A = 78°, B = 41°, c = 108°. 

3. Given B = 115° ; C=50°, a = 70° 20'. 
4. 



Given .4 = 31° 40',- 


C=122°20', 6 = 40° 40'. 


Given A = 108° 12', 


B = 145° 46', c = 126° 32 



162. Case II. Given two sides and their included, angle. 
1. Given b = 138°, c = 116°, .4 = 70° ; find B, C. and a. 
By Napier's Analogies (§ 158), we have 

sin \ (6 + c) _ cot i Jl cos \ (b + c) _ cot -J- ^4 

sin i (6 - c) ~ tan | (i - C')' a " cos | (6 - c) = tan | (.B + C)' 

Whence, tan -i- (B — (7) = sin \ (b — c) esc ^ (b + c) cot -J- ^4, 

+ - 4- 

L tan i (5 + C) = cos | (6 — c) sec £ (6 + c) cot \ A. 

From the data, \ (b - c) = 11°, £ (6 + c) = 127°, £ ^t = 35°. 

og sin \ (b - c) = 9.280599 - 10 log cos J- (6 - c) = 9.991947 • 

og esc i (6 + c) = 0.097651 log sec ± (6 + c) = 0.220537 

log cot i ^1 = 0.154773 log cot \ A = 0.154773 



log tan | (B - C) = 9.533023 - 10 log tan \ (B + C) = 0.367257 

i (B - C) = 18° 50' 24.7". 180° - 1 (B + C) '= 66° 46' 1.2". 

1.(5+0) =113° 13' 58.8". 

Then, B = i (B + C) + £ (5 - C) = 132° 4' 23.5", 

and C = i (5 + (7) - i (5 - C) .= 94° 23' 34.1 ". 

To find a, we have by § 156, 
tan i a ^ sm y(^+^) tan i (6_ c ) ==s i n i (£+ <7) esc ± (5- C) tan 4- (6-c). 

log sin 1 (5 + 0) = 9.963272 - 10 

log esc i (B - C) = 0.490892 
log tan i (b - c) = 9.288652 - 10 

log tan \ a = 9.742816 - 10 

ia = 28°56'51.6", and a = 57° 53' 43.2". 

Note. The triangle is possible for any values of the given elements. 



OBLIQUE SPHERICAL TRIANGLES. 



10; 



EXAMPLES. 

Solve the following spherical triangles : 



'/- 



2. 


Given a = 72°, b = 47°, 


C=33°. 


3. 


Given a = 98°, c — 60°, 


B = 110°. 


4. 


Given b = 70° 40', c = 120°20', 


A = 50°. • 


5. 


Given a = 125° 10', b = 153° 50', 


C = 140° 20'. 



163. Case III. Given the three sides. 

The angles may be calculated by the formulae of § 154. 

If all the angles are to be computed, the tangent formulae are the most 
convenient, since only four different logarithms are required. If but one 
angle is required, the cosine formula will be found to involve the least work. 

The triangle is possible for any values of the data, provided that no 
side is greater than the sum of the other two, and that the sum of the 
skies is less than 360° (§ 131, 1 and 3). 

If all the angles are required, and the tangent formulae .are used, it 
is convenient to modify them as follows. 





By (103), tan £ ^ = ^/ Sm (S 

1 

~~ sin (s — 


— a) sin (s — b) sin (s — c) 
sin s sin 2 (s — a) 




^ |sin (s — a) sin (s — b) sin (s — c) 
a) V sin s 


I 


Denoting Jsin (» - a) sin (» - 
* sin s 


b) sin (s — c) t 7 i 

v by A*, we have 




tan £ A 


k 


sin (s — a) 




In like manner, tan \ B = - — 

sin 




V 


(,s — b) sin (s — c) 




1. Given a = 57°, b = 137°, c 


= 116° ; find A, B, and C. • 




Here, 2s = a + b + c = 310°. 




Whence, s = 155°, s-a = 98 


3 , s-6 = 18°, s-c = 39°. 




log sin (s - a) = 9.995753 - 10 


log k = 9.829330 -10 




log sin - b) = 9.489982 - 10 


log sin (s-b) = 9.489982 - 10 




log sin (s - c) = 9.798872 - 10 


log tan £B=0.339348 




log esc s = 0.374052 


15= 65° 24' 10.4". 




2)19.058659-20 


B = 130° 48' 20.8." 




log A; = 9.829330 -10 


log k = 9.829330 -10 




log sin (s - a) = 9.995753 - 10 


log sin (s-c)= 9.798872 - 10 




log tan -i A = 9.833577 - 10 


log tan£ (7= 0.030458 




i^l = 34°16'52.5". 


- 2 -O = 47°0'27.0". 




.4 = 68° 33' 45.0". 


O=94°0'54.0". 



100 SPHERICAL TRIGONOMETRY. 

EXAMPLES. 

Solve the following spherical triangles : 



2. 


Given a = 38°, 


b = 42°, 


c = 51°. 


3. 


Given a = 101°, 


b = 49°, 


c = 60°. 


4. 


Given a = 126°, 


b = 152°, 


c = 75°. 


5. 


Given a = 62° 20', 


b = 54° 10', 


c = 97° 50 



, find A 

/ 164. Case IV. Given the three angles. 

The sides may be calculated by the formulae of § 155. 

If all the sides are to be computed, the tangent formulae are the most 
convenient, since only four different logarithms are required. If but one 
side is required, the sine formula will be found to involve the least work. 

The triangle is possible for any values of the data, provided that the 
sum of the angles is between 180° and 540° (§ 131, 4), and that each of 
the quantities B+C — A, C + A — B, and A + B—C is between 180° 
and - 180° (§ 134). 

For such values of the angles, S is between 90° and 270°, and each of 
the quantities S - A, S — B, and S—C between 90° and -90°; then, 
cos S is — , while the cosines of S — A, S — B, arid S — C are + (§ 20). 

Hence, the expressions under the radical signs in the formulae are 
essentially positive, and no attention need be paid to the algebraic signs. 

If all the. sides are required, and the tangent formulae are used, it is 
convenient to modify them as follows : 



' , v , cos S cos 2 (S - A) 

By (109), tan£a=-*' 



cos (S - A) cos (S - B) cos (S - C) 



(S-A^jl 



COS ' ' COS o 



cos (S - A) cos (S-B) cos (S - C) 



Denoting J CQS s by K, 

\ cos (S - A) cos (S - B) cos (S - C) 

we have tan \ a ■= K cos (S — A). 

In like manner, tan \ b = Kcos (S — B), and tan \ c = Kcos (S — C). 

1. Given A = 150°, B = 131°, C == 115° ; find a, b, and c. 

Here, 2 S = A + B + C= 396°. 

Whence, S = 198°, S-A = 48°, S-B = 67°, S - C = 83°. 



OBLIQUE SPHERICAL TRIANGLES. 107 

log cos S = 9.978206 - 10 log K= 0.737462 

log sec (S-A) = 0.174489 log cos (S - B) = 9.591878 - 10 

log sec (S-B)= 0.408122 log tan r b = 0.329340 

log sec (S - C) = 0.914106 i 6 = 64° 53' 58.0". 

2 )1.474923 b = 129° 47' 56.0". 

log K= 0.737462 log K = 0.737462 

log cos (S-A) = 9.825511 - 10 log cos (S - C) = 9.085894 - 10 



log tan|« = 0.562973 * 


log tan |c = 9.823356 -10 


la = 74° 42' 4.8". 


i-c = 33°39'23.1". 


a = 149° 24' 9.6". 


c = 67° 18' 46.2". 



Note 1. By § 35, cos 198° = - sin 108° = - cos 18° ; whence, without regard to 
algebraic sign, log cos 198° = log cos 18°. 

2. Given A = 123°, B = 45°, C = 58° ; find a. 



-d /-.™\ • i / cos S cos (S — A) 
By (103), sinifl = \ : — -A — zz — z - 

Here, 2S = A + B+C= 226° ; whence, #=113°, and S - A = - 10 c 

log cos S = 9.591878 - 10 

log cos (S - A) = 9.993351 - 10 
log esc B — 0.150515 

log esc C = 0.071580 

2 JT9.807324 - 20 
log sin i« = 9.903662 -10 

£a = 53° 13' 51.3", and a = 106° 27' 42.6". 

Note 2. By § 29, cos (- 10°) = cos 10°. 

EXAMPLES. 

Solve the following spherical triangles : 

3. Given A = 74°, 5=82°, C=67°. 

4. Given A = 120°, 5 = 130°, (7=140°. 

5. Given ^ = 138° 16', B = 33° 11', O = 36° 53'. 

6. Given A = 91° 10', B = 85° 40', G = 78° 30' ; find b. 

165. Case V. Given tivo sides and the angle opposite to one of them. 
1. Given a = 58°, b = 137°, B = 131° ; find A, C, and c. 

By (85), ^A = *?** or sin A = sin a esc b sin B. 

sin B sin b 



108 



SPHERICAL TRIGONOMETRY. 



log sin a = 9.928420 - 


-10 


log esc b = 0.166217 




Log sin B = 9.877780 - 


-10 



log sin J. = 9.972417 - 10 

.4 = 69° 47' 41.6", or 110° 12' 18.4" (§ 147). 
To find C and c, we have by §§ 156 and 158, 

cot \ C = sin i (b + a) esc \ (6 — a) tan i (5 — ^1), 
and tan i c = sin \ (B + ^4) esc i (i? — ^4) tan i (6 — a). 

Using the first value of A, we have 

1(^ + ^4) = 100° 23' 50.8", and l(B - A)= 30° 36' 9.2". 
Also, i (b + a) = 97° 30', and \ (b - a) = 39° 30'. 



log sin i (b + «) = 9.996269 - 10 
log esc -i- (6 - a) = 0.196489 
log tan | (B - A)= 9.771924 - 10 

log cot | = 9.964682 -10 



log sin i (B + A) = 9.992810 - 1 

log esc i (B -A.)= 0.293214 

log tan | (b - a) = 9.916104 - 10 

log tan ic = 0.202128 

|c= 57° 52' 35.0". 
c= 115° 45' 10.0". 



(7 =94° 39' 15.6". 
Using the second value of A, we have 

!'(£ + -!) =120° 36' 9.2", and ±(B-A) = 10° 23' 50.8". 



log sin £(& + «)= 9.996269 - 10 
log esc i (6 -a) =0.196489 

log tan \ (B -A) = 9.263608 - 10 

log cot -I C 



log sin \(B + A) = 9.934861 - 10 
log esc l(B-A)= 0.743583 
log tan i (b - a) = 9.916104 - 10 



log tan \ c 



0.594548 



\c = 75° 43' 43.6". 
"c = 151° 27 '27.2". 



9.456366 - 10 
i<7= 74° 2' 22.1". 
^0=148° 4' 44.2". 

Thus, the two solutions are : 

1. A= 69° 47' 41.6", C= 94° 39' 15.6", c = 115° 45' 10.0". 

2. A= 110° 12' 18.4", 0=148° 4' 44.2", c = 151° 27' 27.2". 

As in the corresponding case in the solution of oblique plane triangles 
(compare §§ 117 to 120), there may sometimes be two solutions, sometimes 
only one, and sometimes none, in an example under Case V. 

After the two values of A have been obtained, the number of solutions 
may be readily determined by inspection ; for, by § 131, 2, if a is < b, 
A must be < B ; and if a is > b, A must be > B. 

Hence, only those values of A can be retained which are greater or less 
than B according as a is greater or less than b. 



OBLIQUE SPHERICAL TRIANGLES. 109 

Thus, in Ex. 1, a is given < b ; and since both values of A are < B, we 
have two solutions. 

Again, if the data are such as to make log sin A positive, there will be 
no solution corresponding. 

2. Given ct = 58°, c = 116°, = 94° 50'; find A 

T , -, • sin A sin a A ■ n 

In this case, — — — = — — , or sin A = sm a esc c sin C. 
sm C sm c 

log sin a = 9.928420 - 10 
log esc c = 0.046340 
log sin C =9.998453 -10 

log sin .4= 9.973213 -10 

A = 70° 4' 57.1", or 109° 55' 2.9". 

Since a is given < c, only values of A which are < C can be retained ; 
then the only solution is A = 70° 4' 57.1". 

3. Given b = 126°, c = 70°, 5 = 56°; find O. 

t .Li, • sin C sine . ^ f , . ^ 

In this case, = — — , or sm C = sm c esc o sm B. 

sin 5 sm & 

log sin c = 9.972986 - 10 
log esc b = 0.092042 
losr sin B = 9.918574 - 10 



log sin C = 9.983602 - 10 

= 74° 21' 13.8", or 105° 38' 46.2". 

Since both values of C are > B, while c is given < b, there is no solu 
tion. 

EXAMPLES. 

Solve the following spherical triangles : 

4. Given 6= 99° 40', c= 64° 20', B= 95° 40'. 

5. Given a = 40°, b = 118° 20', A= 29° 40'. 

6. Given a = 115° 20', c = 146°20', C=141° 10'. 

7. Given a = 109° 20', c= 82° 1'8", .4 = 107° 40'. 

8. Given b = 108° 30', c= 40° 50', 0= 39° 50'. 

9. Given a = 162° 20', b= 15° 40', J5=125°. 
10. Given a = 55°, c = 138°10', A= 42° 30'. 



110 



SPHERICAL TRIGONOMETRY. 



//. 



166. Case VI. Given two angles and the side opposite to one of them 
1. Given A = 110°, B = 122°, b = 129° ; find a, c, and C. 
sin a sin ^4 



By (85), 



sin b sin 5 



, or sin a = sin A esc JB sin b. 



log sin .4 = 9.972986 -10 
log esc £ = 0.071580 
log sin b = 9.890503 -10 

log sin a = 9.935069 - 10 

a = 59° 26' 37.6", or 120° 33' 22.4" (§ 147). 
To find c and (7, we have by §§. 156 and 158, 

tan \c = sin i (B + A) esc \ (B — A) tan ^(b — a), 
and cot i (7 = sin i (6 + o) esc \(b — a) tan i (5 — A). 

Using the first value of a, we have 

£(& + a)=94°13'18.8", and £(& -a) = 34° 46' 41.2". 
Also, i (5 + A) = 116°, and %(B-A)= 6°. 



log sin i (B + 4) = 9.953660 - 10 
log esc \ (B - A) = 0.980765 
log tani (b - a) = 9.841642 - 10 



log sini (6 + a)= 9.998820 - 10 
log esc 1(6 -a) =0.243821 
log tan i (B - A) = 9.021620 - 10 



logtanic = 0.776067 



log cot 1(7 =9.264261 



*' 



80° 29' 34.8". 
c = 160°59' 9.6". 



10 

hC= 79° 35' 14.1". 
C= 159° 10' 28.2"." 



Using the second value of a, we have 

a)= 124° 46' 41.2", and i (b - a) = 4° 13' 18.8". 



i(P 



log sin %(B + A) = 9.953660 - 10 

log esc 1 (B - ^) = 0.980765 

log tan £ (6 - a) = 8.868171 - 10 



10 



log tan|c = 9.802596 



10 

c = 32° 24' 17.8". 
c = 64° 48' 35.6". 



log sin i (b + a) = 9.914537 
log esc | (6 - a) = 1.13300ft 

log tan i (B -A)= 9.021620 - 10 

log cot 1(7 = 0.069166 

1(7= 40° 27' 24.1". 
C =80° 54' 48.2". 



Thus the two solutions are : 

1. a= 59° 26' 37.6", c = 160°59 r 9.6", O = 159° 10' 28.2". 

2. a = 120° 33' 22.4", c= 64° 48' 35.6", (7= 80° 54' 48.2". 

In examples in Case VI., as well as in Case V., there may sometimes 
be two solutions, sometimes only one, and sometimes none. 



OBLIQUE SPHERICAL TRIANGLES. 



Ill 



As in Case V., only those values of a can be retained which are greater or 
less than b according as A is greater or less than B. 

Also, if log sin a is positive, the triangle is impossible. 

EXAMPLES. 
Solve the following spherical triangles : 

2. Given B = 116°, C= 80°, c= 84°. 

3. Given A = 132°, £=140°, 6 = 127°. 

4. Given A= 62°, (7= 101° 58' 24", a= 64° 30'. 

5. Given A = 133° 50', B= 66° 30', a= 81° 10'. 

6. Given B= 22° 20', C'=146°40', c - 138° 20'. 

7. Given ^1= 61° 40', C= 140° 20', c = 150°20'. 

8. Given 5= 73°, C= 81° 20', b = 122° 40'. 



APPLICATIONS. 



167. In problems concerning navigation, the earth may be regarded as 
a sphere. 

The shortest distance between any two points on the surface is the arc 
of a great circle which joins them ; and the angles between this arc and 
the meridians of the points determine the beanngs of the points from each 
other. 




Thus, if Q and Q' are the points, and PQ and PQ' their meridians, the 
angle PQQ' determines the bearing of Q' from Q, and the angle PQ'Q 
determines the bearing of Q from Q'. 

If the latitudes and longitudes of Q and Q' are known, the arc QQ 1 
and the angles PQQ' and PQ'Q may be determined by the solution of a 
spherical triangle. 

For if EE' is the equator, and PG the meridian of Greenwich, we have 

Z QPQ' =Z Q'PG -Z QPG = longitude Q' - longitude Q. 



112 SPHERICAL TRIGONOMETRY. 

Also, PQ =PE - QE = 90° - latitude Q, 

and PQ' = PE' + Q'E' = 90° + latitude Q'. 

Thus, in tlie spherical triangle PQQ', two sides and their included 
angle are known, and the remaining elements may be computed. 

When QQ' has been found in degrees, its length in miles may be 
calculated by finding the ratio of its arc to 360°, and multiplying the 
result by the length of the circumference of a great circle ; in the following- 
problems, the radius of the earth is taken as 3956 miles. 

EXAM PLES. 

1. Boston lies in lat. 42° 21' X., Ion. 71° 4' W. ; and the latitude of 
Greenwich is 51° 29' X. Find the shortest distance in miles between the 
places, and the bearing of each place from the other. 

2. Calcutta lies in lat. 22°33'X., Ion. 88° 19' E. ; and Valparaiso in 
lat. 33° 2' S., Ion. 71° 42' W. Find the shortest distance in miles between 
the places, and the bearing of each place from the other. 

3. Sandy Hook lies in lat. 40° 28' X., Ion. 74° 1' W. ; and Queenstown 
in lat. 51° 50' X., Ion. 8° 19' W. In what latitude does a great circle 
course from Sandy Hook to Queenstown cross the meridian of 50° W. '.' 

168. The Astronomical Triangle. 

z 

E 
M 




Let O be the position of an observer on the surface of the earth : P the 
celestial north-pole ; Z the zenith. 

The great circle EE', having P for its pole, is called the celestial equa- 
tor; and the great circle HH\ having Z for its pole, is called the horizon. 

Let S be the position of a star ; PSM a meridian passing through S : 
ZSN a quadrant of a great circle passing through Z and S. 

The arc SM is called the declination of the star; and is called declina- 
tion north or south, according as the star is north or south of the celestial 
equator. 

The angle SPZ is called the hour-angle of the star; the arc SN is 
called its altitude; the angle PZS, its bearing or azimuth. 



OBLIQUE SPHERICAL TRIANGLES. 113 

The arc EZ is the latitude of the observer. 

The spherical triangle SPZ is called the Astronomical Triangle; its 
sides have the following values : 

SP = PM— SM= 90° - the decimation of the star ; 

SZ = ZN - SN = 90° - the altitude of the star ; 

PZ=EP -EZ = 90° - the latitude of the observer. 

Its angle SPZ is the hour-angle of the star, and its angle SZP the 
azimuth. 

If any three of these five elements are known, the solution of a spheri- 
cal triangle serves to determine the other two. 

169. To Determine the Hour of the Day. 

If the altitude and declination of the sun are known, and the latitude 
of the observer, the three sides of the triangle SPZ are known, and the 
hour-angle SPZ may be computed. 

If 24 hours be multiplied by the ratio of this angle to 360°, we obtain 
the time required for the sun to move from S to the meridian EP. 

If this time be subtracted from 12 o'clock, if the observation is made 
in the morning, or added, if made in the afternoon, we obtain the hour of 
the day at the time and place of observation. 

If the Greenwich time of the observation be noted on a chronometer, 
the difference between this and the local time as calculated above serves 
to determine the longitude of the place of observation. 

In reducing time to longitude, it should be remembered that 24 hours 
of time correspond to 360° of longitude ; that is, one hour of time corre- 
sponds to 15° of longitude, one minute to 15', and one second to 15". 

EXAMPLES. 

170. 1. A mariner observes the altitude of the sun to be 14° 18', its 
declination being 18°36'N. If the latitude of the vessel is 50°13'N., 
and the observation is made in the morning, find the hour of the day. 
If the observation is taken at 9 a.m., Greenwich time, what is the longi- 
tude of the vessel ? 

2. What will be the altitude of the sun at 4 p.m. in San Francisco, 
lat. 37°48'N., its declination being 12° S. ? 

3. What will be the bearing of the sun at 9.30 a.m. in Melbourne, lat. 
37° 49' S., if its declination is 6° S. ? 

4. At what hour will the sun rise in Boston, lat. 42°21'N., if its 
declination is 15° N. ? 

Note. At sunrise the sun's altitude is 0, so that the arc SZ becomes 90°. 



114 TRIGONOMETRY. 

FORMULAE. 

PLANE TRIGONOMETRY. 

§29. sin(— A)= — sin A. tan (—A)= —tan A. sec(— A)= sec -4.] 
cos(— A)= cos A cot ( — A)= — cot A. esc (—A)= — esc A. J 



C 1 ) 



§30. sin (90° + A) = cos A. cot (90° + A) = - tan A. j 

cos (90° + ^4) = - sin A. sec (90° + A) = - esc A [ (2) 

tan (90° +i) = - cot A. esc (90° + A) = sec A. J 

§ 36. sin a? = tan a? == sec a? 



esc x cot a; cos x 

eosx = cot x = esc x = ■ 



sec a; tan a; sin a; 



(3) 



§37. tanx = ^^. ( 4 ) 

COS X 

§ 38. cot x = ^£L ( 5 ) 

sm a; 

§ 39. sin 2 x + cos 2 a? = 1. (6) 

§ 40. sec 2 a? = 1 + tan 2 x. (7) esc 2 a? = 1 + cot 2 x. (8) 

§ 41 sin (x -f- ?/) — sin a; cos ?/ + cos a; sin ?/. (9) 

cos (x + ?/) = cos a; cos y — sin a; sin y. (10) 

§ 43. sin (x — y) = sin a? cos ?/ — cos x sin ?/. (11) 

cos (x — y) = cos x cos ?/ + sin x sin ?/. (12) 

a a a / , \ tanx+tany /n _ N . , . . cotxeotv— 1 /-,«-\ 

§44. tan(x+?/) = — , , y • 3 ) cot(x+?/) = — *- (15) 

1— tan a; tan?/ cot?/+cotx 

. z N tan x— tan?/ ,,_>. m . , N cotxcotv+1 ,,-k 

tan (x — ?/) = — • (14) cot (x— y)= 2 (16) 

v J 1+tanxtan?/ v ' v ' cot?/-cotx v ' 

§ 45. sin x + sin y — 2 sin \{x + y) cos ^ (a; — y). (17) 

sin a; — sin y = 2 cos J (x -f- ?/) sin \ (x — y). (18) 

cos x + cos 2/ = 2 cos |- (a; + y) cos J (a; — y). (19) 

cos x — cos y = — 2 sin i (a; + y) sin -J- (x — y). (20) 

sin x + sin y tan *- (a; + y) 

sm a; — sm ?/ tan \{x — y) x 7 

§ 47. sin (a; + y) sin (a: — y)= sin 2 a; — sin 2 y. (22) 

sin (a; + y) sin (a; — y) = cos 2 ?/ — cos 2 x. (23) 

cos (a; + y) cos (x — y) = cos 2 x — sin 2 y = cos 2 ?/ — sin 2 x. (24) 



FORMULAE. Ho 

§ 48. sin 2 x = 2 sin x cos x. (25) cos 2 x — 2 cos 2 x — 1. (28) 

cos 2 # = cos 2 a? — sin 2 x. (26) tan 2x = (29 ) 

v J 1 — tan 2 a; 

cos2a- = l-2sin 2 a;. (27) C ot2a;=^* ; — • (30) 
v ; 2coto; 

§49. 2sin 2 i^ = l-coso;. (31) tan^o? = 1 ~ cosx . (33) 

y sinx 

2cos 2 i.T = lH-coso:. (32) cot 4 a; = * + C0S * (34) 
2 v J z sin a; 

§50 sin 3 a? = 3 sin x — 4 sin 8 x. (35) cos 3 a; = 4 cos 3 # — 3 cos x. (36) 

, o 3 tan x — tan 3 x /0 _s 

tan 3 a; = — — — (37) 

1—3 tairaj 

§ 105. 4 K = c 2 sin 2 A (38) 2 A" = or tan #. (42) 

4 AT = c 2 sin 2 I?. (39) 2 A" = U 1 tan .1. (43) 



2if=a 2 cotA - (40) 2A"=«V(c + a)(c-a). (44) 



2K=b 2 cot 5. (41) 2 K = b V(c + b) (c - &). (45) 

2K=ab. (46) 

§ 107. a : b = sin yl : sin B. (47) 

6 : c = sin JB : sin C. (48) 

c : a = sin C y : sin A. (49) 

S10 o a + & tan^(A+ g) , 5Q x 

U08 ' ^6 = tanX(^--5)* • (50) 



fr + c_tanj-(J3+C) 
6-c~"tan^(5- C) 

c + a ta.n±(C+A) 



(51) 



(52) 



c — a tan ±(C — A) 

§ 109. a 2 = b 2 +c 2 -2bc cos A (53) 

6 2 = c 2 + a 2 - 2 ca cos 5. (54) 

o 2 = a 2 + & 2 - 2 a& cos C. (55) 

011A , £r + c 2 — a 2 •__. -o c 2 + a 2 — b 2 /e _ x 

§ 110. cos A = — — (56) cos Z? = — — (57) 

2 be K ) 2ca K J 

si a 2 -\-b 2 — c 2 /no s 

cos C = — -*- (58) 



2ab 


/? 


- &)(« - 


-•) 


» 


/cr- 


-«)(•- 


-a) 



§111. sin|yl=^ ^ ^« (59) 

siiUB=^/— 



ca 



116 TRIGONOMETRY. 



sin 



" * be 





sin 


A 


sin a 
sine 


(76) 




sin B ■■ 


_ sin ft 
sin c 




cos 


A 


tan b 
tanc 


(77) 




cos B ■■ 


_ tan a 
tan c 


§137. 


tan^l = 


_ tana 
sin b 


(80) 




tsmB 


_ tan ft 
sin« 


§ 138. 


sin 


A-. 


cosB 
cos 6 


(82) 




sin B ■■ 


cos A 

cos a 


§ 139. 








cos c = cot A cot B. 






§ 151. 








sin .4 


sin a 






sin B 


sin b 












sini? _ 


sin& 














sin C 


sine 














sin A 


sin a 














sinC - 


sin c 







(61) 



C0S ^=M 6c • ( 62 ) 

cosiB ^ s J^l3. (63) 



ta i i= JSO. (65) 

2 * s (s — a) 

2 * s (s — ft) 

taniO=V^fc^- (67) 

2 * s (s — c) 

§112. 2 /iT=ftc sin A (68) 2K=i a 2 smBsmC m ^ 

sin .1 

2/f=casm5. (69) 2 g== ^ sin C sin ^ ( 

v ; sinJ5 v ; 

2^r=a6sinO. (70) 2 7T = ^ sin ^ sin B . (73) 

sm C 

J5T = Vs (s — a)(s — ft)(s — c). (74) 

SPHERICAL TRIGONOMETRY. 
§ 135. cos c = cos a cos ft. (75) 

(78) 

(79) 

(81) 



<83i 
(84) 

(85) 

(86) 

(87) 



FORMULAE. 117 

§ 152. cos a = cos b cos c -f- sin & sin c cos A. (88) 

cos 6 = cos c cos a -f- sin c sin a cos B. (89 ) 

cos c = cos a cos 6 -f- sin a sin 6 cos C. (90) 

§ 153. cos A — — cos _B cos C -I- sin 5 sin C cos a. (91) 

cos B = — cos (7 cos ^L + sin C sin ^4 cos b. (92) 

cos C = — cos A cos B + sin ^4 sin B cos c. (93) 



_ ~ , • , , /sin (s — b) sin (s — c) , , 

§ 154. sin I A = a! v . / . v -. (94) 

* sm b sm c v y 



. , ._ Ism (s — c) sm (s — a ) . N 

smlJ3 = \ i — . ; . -• (95) 

* sm c sm a x ' 



/sin (s — a) sin (s — b) 
sin a sin 6 



sm i C = A ^ — . — L^_A ( 96 ) 

11 Sill tl. Sill h x J 



. . /sin s sin (s — a) , . 

COS i ^1 = \ r-^-A ^- (97) 

* sm b sm c 



, ^ /sm s sm (s — b) , N 

cos -i- £ == \/ : ^ 1- . (98) 

* sin c sm a v y 



sm c sm a 



, ~ /sin s sin (s — c) , . 

cos I C = a] -. ^—r— • (99) 

L * sm a sm 6 v 7 



. . /sin (s — b) sin (s — c) , 

tan 4 .4 = a\ v J—, — K —— J -> (100 ; 

z « sm .s- sm (a — rn 



, „ /sin (s — c) sin (s —a) , . 

tan£J5 = V - • / in • ( 101 ) 

* sin s sm (s — 6) 

. _, /sin (s — a) sin (s — 6) ,^ . 

tan i C = \/ ^ / , v N — - • (102) 

* sm s sm (s — c) 



o, **>*> -i / cos S cos (S — ^4) /,«o\ 

§ 155. sm i a = V . _/. „ ; - (103) 

^ if sin H sm f ; 



sin B sin C 

/ p.ns 

sin ^ 6 



/ _ cos 8 cos (£ - B) ^ 1Q4) 

\ sin (I sin /I 



cos # cos (S - C) , M 

sm$c=A] -. — . \ p ' - (105) 

2 * smi sm B 

0OSl« = #^3i#^. (106) 



sin 5 sin C 



118 TRIGONOMETRY. 



/cos (S — C) cos (S - A) , 

b = yj i — . ) . v /. (107 ) 

* sin ( 1 sin A 



x sin Osin^l 



/cos (S-A) cos (S-B) 
cos i c = ^/ ^^^ri— (108) 



sin ^4 sin 5 



Vcos S cos (# — -4) 
715 ^ /cr m ' ( 109) 
cos (aS — i2) cos (# — (?) 

I cos S cos (S - 5) 

tan^6=-v/ -^ -^ — /ct ; . • (110) 

2 * cos (# — C) cos (/&' — ^4) 



/ cos £ cos (S-C) 

2 * cos (£ — ^4) cos (aS — 5) 

_ iW sin f /- ^1 + -B) tan^c , 

sm -J- (J. — 5) tan £ (a — 6) 



8 -, k 7 cos -J- (^4 + jB) _ tan \ c 

' cos I (A - B) ~ tan i (a" + &)' . 

g -j ,.£ sin i (a + b) _ cot -^ C 

sin l( a -b)~ tan| (i - B) 

cos \ (a + 6) _ cot i (7 
cos i (a — b) ~ tan i ( A + 5) 



(113) 
(114) 

(115) 



ANSWERS. 









§ 56 ; page 30. 


12. 


85° 56' 37.32". 




14. 95° 29' 34.8". 


13. 


14° 19' 26.22". 




15. 20° 27' 2.52". 
§ 63 ; page 39. 


3. 


n-rr, 2mr ± — 
o 




7. 


4. 


(2n + l)|, mr 


+ (- 


1)*^. 8. 

' 6 


5. 


(2» + l)| »*■ 


±r 


9. 



16. 130° 55' 5.952" 



u-k ± 



)l7T ± 



±tan-*f±Vn 



6. 


4 




§ 


76; 


10. sin- 
page 44. 


iVS-1 

2 




2 


1.5441. 


6. 


2.1003. 




10. 2.5104. 


14. 


3.4192. 


3. 


1.6990. 


7. 


2.2922. 




11. 2.5774. 


15. 


3.7814. 


4. 


1.6232. 


8. 


2.3892. 




12. 2.9421. 


16. 


4.0794. 


5. 


1.8751. 


9. 


2.3222. 

§ 


78, 


13. 2.8363. 
page 44. 


17. 


4.2006. 


2. 


0.5229. 


5. 


1.1549. 




8. 0.2831. 


11. 


1.4592. 


3. 


0.2431. 


6. 


0.2589. 




9. 0.7939. 


12. 


1.3468. 


4. 


1.6532. 


7. 


2.3522. 

§• 


81; 


10. 2.1303. 
page 45. 


13. 


2.0424. 


3. 


3.3397. 


8. 


0.5663. 




13. 0.6171. 


19. 


0.8752. 


4. 


1.7475. 


9. 


0.0430. 




14. 0.2918. 


20. 


0.0794. 


5. 


0.6338. 


10. 


0.1165. 




16. 0.0495. 


21. 


0.4248. 


6. 


8.6826. 


11. 


0.0939. 




17. 0.0365. 


22. 


0.1051. 


7. 


1.0460. 


12. 


0.5440. 




18. 0.7007. 


23. 


0.0406. 



119 



120 









ANSWERS. 












§ 85 ; page 47. 






2. 


0.5562. 


5. 


8.9912 


-10. 8. 8.5932- 


-10. 


11. 2.3064. 


3. 


1.0491. 


6. 


7.5353 


-10. 9. 6.6074- 


-10. 


12. 0.1151. 


4. 


9.9242 - 10. 


7. 


3.4592. 


10. 9.2885- 


-10. 


13. 0.7782. 








§ 86; page 47. 






4. 


0.011739. 




10. 


4.942550 - 10. 


18. 


186.334. 


5. 


2.527511. 




11. 


5.863566. 


19. 


.00223905. 


6. 


6.780210 - 10. 




12. 


5.640409 - 10. 


20. 


.0000100006. 


7. 


4.812917. 




15. 


6.61005. 


21. 


9776.67. 


8. 


3.960116. 




16. 


55606.5. 


22. 


467929. 


9. 


7.013152 - 10. 




17. 
24. 

§ 91 


.0110890. 
.00000130514. 

; pages 50, 51. 


23. 


.000342770. 


1. 


1897.85. 




17. 


244.004. 


35. 


.695490. 


2. 


- 193315. 




18. 


.00279116. 


36. 


.542699. 


3. 


.309170. 




19. 


.000000237177. 


37. 


- 36.0189. 


4. 


.00110375. 




20. 


2.23607. 


38. 


- 11.1122. 


5. 


6.36103. 




21. 


1.14870. 


39. 


.943241. 


6. 


.0301742. 




22. 


- 1.22028. 


40. 


2.62762. 


7. 


31.2004. 




23. 


1.77828. 


41. 


2.53217. 


8. 


- .132693. 




24. 


.668289. 


42. 


- 1.79616. 


9. 


.126965. 




25. 


.645831. 


43. 


1.03242. 


10. 


.0235770. 




26. 


.137751. 


44. 


.298557. 


11. 


- 1.16493. 




27. 


- .370134. 


45. 


.0448607. 


12. 


- .00256105. 




30. 


13.8289. 


46. 


.794509. 


13. 


3692.77. 




31. 


2.48722. 


47. 


1.80492. 


14. 


.277996. 




32. 


1.05557. 


48. 


179.596. 


15. 


- 15896.0. 




33. 


.0000214279. 


49. 


1.88270. 


16. 


.0316228. 




34. 


.00710469. 


50. 


.000193152. 




51. - 


.0995935. 


52. 1.34384. 








§ 92 ; page 52. 






3. 


x = .2831 + . 
£ = -2.173+. 






8. ,-j. 


3 log a 


4. 


log n 


— 2 log m 


5. 
6. 


x =1.155+. 
a = -.1765 + . 






»■ .-i 






7. 


x _ 51ogc 




b 


10. x = l 


or — 


5. 


log a — 2 log 











ANSWERS 




121 








§ 93 ; page 52. 






2. 


3.7004+. 3. 


- .06546 + . 4. 


— 


6.059 + . 5. 3.326 + . 




6. 


-.4601 + . 7. 


.3494 + . 9. 4. 


10. 


5 . 11. - 1 - 12. 6 - 

3 3 








§ 94 ; page 53. 






I. 


9.345950 - 10. 


7. 0.302190. 




13. 27° 31' 50.5'. 




2. 


0.376890. 


8. 0.153906. 




14. 8° 41' 32.7". 




3. 


9.932630 - 10. 


9. 0.002256. 




15. 75° 45' 9.8". 




4. 


9.865995 - 10. 


10. 59° 15' 26.4". 


16. 49° 38' 57.1". 




5. 


9.243533 - 10. 


11. 33° 0' 16.1". 


17. 23° 26' 30.9". 




6. 


9.163433 - 10. 


12. 81° V 37.9". 
§ 95 ; page 53. 






1. 


.68573. 


4. .69518. 




7. 51° 36' 42.9". 




2 


.25232. 


5. .92163. 




8. 15° 28' 22.5". 




3. 


.06344. 


6. .86962. 




9. 66° 14' 34.3". 








10. 29° 9' 13.8". 










§ 96 ; page ! 


53. 






1. 


8.338076 - 10. 


3. 1.369926. 




5. 0° 24' 53.79". 




2 


8.810945-10. 


4. 0° 58' 51.06". 


6. 1° 37' 41.93". 








§ 102 ; pages 56 to 58. 


1. 


a- 


= 1.8117, b = 6 


.7615. 14. 


a = 


= 176.533, c= 191.993. 


2 


b-- 


= 11.7793, c = 12.7965. 15. 


a = 


= 20455.6, c = 21405.6. 


3. 


a- 


= 16.7820, c = 26.1081. 16. 


a = 


: 2.40989, b = .812578. 


4. 


A-- 


= 34° 22' 7.1", b = 


: .511764. 17. 


A = 


: 19°31'57.2",c=.000505172 


5. 


A-- 


= 33° 8' 56.3", c = 


499.252. 18. 


b = 


= 77.6330, c = 91.2952. 


6. 


b-- 


= 10.3547, c = 13.1404. 19. 


A = 


: 32° 10' 16.5", a = 388.471. 


7. 


a - 


= .0036235, b = .013523. 20. 


b = 


-- 644.109, c = 650.272. 


8. 


A-- 


= 39° 49' 24.6", a 


= 48.8645. 21. 


a = 


,34308.0, 6 = 23381.6. 


9. 


a - 


= 148.407, c = 948.680. 22. 


b = 


: 4.48174, c = 8.5085. 


10. 


A-- 


= 49° 53' 54.9", c = 


= 4.46330. 23. 


A = 


: 39° 21' 54.1", 6 = 121.240. 


11. 


6 = 


= .000336374, c = 


.00336715. 24. 


a = 


: .00247181, c=. 00360016. 


12. 


a - 


= 3821.55, 6 = 3641.34. 25. 


a = 


16001.6, c = 85725.1. 


13. 


.1 = 


-35° 53' 55.2", 6: 


= 731.237. 26. 


a = 


3624500, 6 = 8821960. 



122 



ANSWERS. 



27. A = 76° 33' 49.0", a = 24234.4. 

28. a = 207302, 6 = 421170. 



29. a = .507624, c = .525355. 

30. .4 = 60° 14' 12.9", c = 774.563. 



36. a = 4925.31. 

37. 20.573. 

38. 83.271 ft. 

39. 31° 47' 24.5". 

40. 36° 37' 58.0". 



41. 99.4565 mi. 

42. 10.2352. 

43. 19° 49' 46.7". 

44. 365.64 ft. 

45. 56° 18' 35.7". 



31. c = 252.103. 

32. a = 1.73561. 

33. c = 122748. 

34. .4 = 47° 42' 47.8". 

35. « = . 344647. 

46. 25.2230 ini., 30.0750 mi. 48. 14.4853, 15.6787. 

47. 21.6514. 49. 517.51ft. 

50. 17.2624. 51. 420.867 ft. 52. 437.605. 

53. 10.392. 54. 482.1 ft. 

55. Rate, 6.79668 miles an hour; Rearing, K 63° 8' 28.5" W. 



2. £ = 89° 59' 42.8". 

3. B =89° 23' 22.6". 

4. ^4 = 89° 59' 37.2". 



§ 104 ; page 60. 

5. 5 = 89° 59' 59.0". 

6. A = 89° 43' 13.6". 



2. 6.9066. 

3. .151079. 

4. 5699.7. 



§ 106; page 61. 

5. .089433. 

6. 8130.9. 

7. .0067825. 



8. 2.18876. 

9. 107.762. 
10. .0487840. 



§ 114 ; page 67. 



2. 6 = 283.331, c = 267.677. 7. 

3. a = .340132, c=. 986084. 8. 

4. a = 29.0595, 6 = 18.3742. 9. 

5. a = .0313440, c = .0498733. 10. 

6. 6 = 5.76721, c = 2.16917. 



a = 5058.5, 6 = 3683.53. 
a = .299674, 6 = .731538. 
a = 4.01036, c = 3.55195. 
6 = 56719.9, c = 23073.5. 



§ 115 ; pages 68, 69. 
2. ^4 = 118° 17' 57.4", 6=44.7274. 7. (7=63° 48' 28.1", 



3. A= 60° 44' 39.5", c=965.282. 

4. C= 63° 49' 9.3", a=4.48237. 

5. B= 28° 43' 49.0", c= 1.44246. 

6. 5=145° 35' 24.7", a = 1045.74. 



6 = 13.7387. 
c=85.3596. 



8. ^4 = 67° 55' 16.9", 

9. (7=46° 13' 20.9", 
10. C=134°36'27.4", 6 = 27335.0. 



9. C=46°13'20.9", « = .07595S8. 



ANSWERS. 123 

§ 116; page 70. 

3. ^ = 28° 57' 18.0", B = 46° 34' 2.8", 0= 104° 28 '39.0". 

4. .4 = 44° 24' 54.8", B= 78° 27' 47.0", C= 57° 7' 17.6". 

5. A = 71° 47' 24.4", B = 58° 45' 5.4", 0= 49° 27' 30.0". 

6. ^. = 74° 40' 16.4", B= 47° 46' 39.0", C= 57° 33' 4.8". 

7. .4 = 59° 19' 11.8", B= 68° 34' 7.6", (7= 52° 6' 40.6". 

8. A -45° 11 '46.6", B = 101° 22' 17.8", C = 33°25'56.4". 

9. ^ = 71° 33' 49.2". 10. B = 30° 47' 22.8". 11. C = 25° 56' 54.2". 

§ 121 ; pages 73, 74. 

1. 5= 32° 36' 9.4", c = 6.62085. 

2. #!= 31° 57' 47.8", «!=: 120.313; 
#, = 148° 2' 12.2", a 2 = 11.3800. 

3. C= 23° 33' 18.2", a = .183882. 

4. J. = 34° 29' 48.2", 6 = 7.12905. 

5. Impossible. 

6. Impossible. 

7. B = 48° 34' 38.4", a = 76.0172. 

8. C=90°, 6 = 5.51109. 

9. Ci= 46° 18' 35.5", ^ = 6.94575; 
C 2 = 133° 41' 24.5", a, = .699906, 

10. A= 25° 32' 50.9", c = 278.193. 

11. Impossible. 





> 


12. 


C= 14° 4' 7.7", 6 = 1.43516. 






13. 


B = 90°, c = 137.872. 






14. 


A x = 70° 12' 46.7", ^ = .287904 
.L = 109°47'13.3", 6 2 = . 104539 






15. 


B= 45° 38' 30.2", a = 16214.3 
§ 122 ; pages 74, 75. 


2. 


197.656. 




5. 165917. 8. .078614. 


3. 


14.9812. 




6. 2878.31. 9. 860.006. 


4. 


16.6843. 




7. 1.30108. 10. .0448746. 



11. 4000.81. 

12. .000329015. 

13. 25.6249. 

§ 123; pages 75, 76. 

1. Height, 153.629 ft. ; distances, 117.246 ft, 217.246 ft. 

2. AD = 44.9525. 4. 47° 52' 2.1". 6. 56.6547,49.3482. 

3. 29799.9 sq. rd. 5. 247.741 ft. 7. 32.5255 mi. 



124 ANSWERS. 

8. Two angles, 74° 12' 20.0", 58° 23' 48.0"; third side, .430133. 

9. K 47° 32' 33.1" W. 10. 9.8995 mi., 19.1244 mi. 

11. One angle, 101° 13' 45.8"; diagonal, 136.187. 12. 297.954 ft. 

13. Sides, 26.5604, 90.5154 ; one angle, 119° 5' 14.6". 

14. 91.6364 ft., 33.8973 ft. 15. 17.64934,8.77461, 
16. 1113.34 ft. 17. Diagonal, 52.9024; side, 41.9506. 
18. 247.998 ft. 19. AD = 88.1534, A = 56° 1' 10.7". 

20. 1569.948 sq. rd. 

§ 126 ; page 79. 

2. 2.11491, -1.86081, -.254102. 4. .47761, - 6.1364, - .34120. 

3. 2.14510, .523978, - 2.66907. 5. 3.49086, - .83425, .343379. 



§ 148 ; page 93. 

5. A= 36° 58' 50.0", B = 63° 42' 34.0", 6= 42° 34' 54.4". 

6. a= 27° 49' 17.9", b = 42° 29' 21.8", c = 49° 17' 42.4". 

7. B = 68° 37' 18.1", 6 = 44° 56' 46.7", c= 49° 20' 41.8"; 
or, B = 111° 22' 41.9", 6 = 135° 3' 13.3", c = 130° 39' 18.2". 

8. A= 68° 10' 4.4", b = 163° 42' 32.1", c = 141° 50' 15.2". 

9. A= 15° 34' 32.3", B= 94° 14' 40.0", c = 105° 26' 27.5". 

10. a = 170° 13' 25.6", B= 78° 34' 3.4", b= 40° 1' 8.6". 

11. A= 21° 11' 12.7", a= 19° 50' 30.4", c = 69° 54' 41.6"; 
or, ^t = 158° 48' 47.3", a = 160° 9' 29.6", c = 110° 5' 18.4". 

12. A= 82° 8' 19.3", 

13. A = 122° 34' 33.5", 

14. ^ = 153° 10' 2.8", 

15. .4 = 165° 50' 26.0", 

16. a = 112° 16' 49.7", 

17. A= 55° 58' 5.5", 

18. a= 54° 0'24.8", 

19. a= 41° 29' 25.7", 

20. a = 152° 35' 19.0", 
21.-4= 20° 3' 21.5", 
or, .4 = 159° 56' 38.5", 

22. a = 110° 57' 15.6", 

23. A = 111° 53' 21.2", 



a= 73° 38' 54.4", 


b= 28° 4' 23.5". 


a = 132° 24' 39.6", 


B= 52° 58' 9.5". 


B =115° 25' 2.8", 


c= 20° 2' 40.3". 


b = 139° 10' 11.5", 


c= 41° 42' 23.4". 


b = 145° 51' 35.5", 


c= 71° 42' 41.1". 


5= 34° 41' 20.4", 


c= 12° 39' 44.7". 


B= 84° 43' 10.5", 


c= 86° 10' 32.3". 


b = 133° 39' 29 8", 


c = 121° 8' 21.5". 


JB = 108° 7' 8.6", 


6 = 125° 24' 13.7". 


a= 14° 58' 21.1", 


c = 131° 7' 4.9"; 


a = 165° 1'38.9", 


c= 48° 52' 55.1". 


J5= 165° 10' 31.9", 


c= 69° 41' 7.1". 


£ = 115° 40' 6.8", 


6 = 117° 49' 41.2". 



ANSWERS. 125 

24. A = 165° 3' 57.9", a = 168° 8' 48.3", 6= 51° 53' 53.3". 

25. 5= 22° 13' 3.9", b = 20° 34' 38.3", c = 111 38' 31.1"; 
or, B = 157° 46' 56.1", 6 = 159° 25' 21.7", c = 68° 21' 28.9". 

26. A = 64° 30' 52.0", a = 38° 32' 30.5", 5= 146° 37' 27.3". 

§ 149 ; page 94. 

2. a = 103° 25' 57.4", B = 157° 31' 44.4", 0= 119° 19' 11.3". 

3. o= 57° 43' 57.2", b = 129° 56' 31.7", = 5S° 4' 55.6". 

4. .4= 19° 56' 45.0", B = 141° 38' 20.3", 6 = 113° 18' 58.3". 

5. ^4 = 44° 41' 15.9", a = 51° 37' 1.9", 6 = 60° 51' 3.4". 

6. B= 80° 27' 25.7", 6= 80° 46' 54.3", = 87° 31' 12.5"; 
or, B= 99° 32' 34.3", 6 = 99° 13' 5.7", = 92° 28' 47.5". 

7. 4= 67° 11' 45.0", B= 80° 58' 16.5", = 93° 29' 13.4". 

§ 150 ; page 95. 

2. a= 69° 55' 43.2", 0= 159° 59' 40.6". 

3. A = 120° 41' 19.6", c = 30° 14' 37.4". 

4. .4 = 140° 35' 4.5", 0=145° 11' 50.4". 

5. 0=148° 19' 24.8", c= 80° 47' 39.8". 

§ 161 ; page 104. 

2. a= 95° 37' 51.0", 6= 41° 52' 22.2", = 110° 48' 24.0". 

3. b= 98° 30' 32.4", c= 56° 42' 47.0", A= 59° 38' 53.2". 

4. c = 64° 19' 27.8", a = 34° 3' 11.8", B= 37° 39' 27.2". 

5. 6 = 146°25' 1.4", a= 69° 4' 38.2", 0= 125° 11 '41.8". 

§ 162 ; page 105. 

2. A = 121° 32' 41.3", B= 40° 56' 48.5", c= 37° 25' 48.8". 

3. A= 86° 59' 48.8", = 60° 50' 54.8", b = 111° 16' 42.4". 

4. = 134° 57' 31.3", B= 50° 40' 48.3", a = 69° 7' 34.6". 

5. 5 = 163° 8' 48.4", A = 147° 29' 24.2", c= 76° 8' 49.0". 

§ 163 ; page 106. 

2. A= 51° 58' 28.0", B= 58° 53' 13.2", 0= 83° 54' 31.6". 

3. .4 = 142° 32' 37.8", B= 27° 52' 36.0", 0= 32° 26' 52.8". 

4. .4 = 142° 23' 44.0", J5 = 159° 15' 41.6", 0=133° 14' 4.2". 

5. A= 47° 21' 11.8". 



126 ANSWEBS. 

§ 164 ; page 107. 



c= 63° 12' 24.6". 
c = 132° 5' 10.0". 
c= 61° 14! 18.2". 



a = 100° 42' 23.4". 
c = 153° 29' 39.8"; 
c = 90° 8' 51.4". 

6 = 114° 47' 47.5". 
a = 131° 16' 32.2"; 
a= 95° 48' 41.8". 

6= 96° 33' 16.2". 



A = 79° 18' 29.0". 
C=164° 6' 8.4"; 
C= 128° 22' 54.8". 
6= 66° 29' 37.6". 



6. 6= 27° 22' 7.6", a = 117° 9' 5.2", A = 47° 20' 57.2". 

7. a= 43° 2' 23.6", 5 = 129° 9' 46.0", B = 89° 23' 51.8"; 
or, a -136° 57' 36.4", b = 20° 34' 54.2", 5= 26° 57' 36.4". 

8. Impossible. 

§ 167 ; page 112. 

1. Distance, 3275.20 mi. ; bearing of Boston from Greenwich, X. 71 c 
38' 53.7" W. ; of Greenwich, from Boston, N. 53° 6' 31.9" E. 

2. Distance, 11012.9 mi. ; bearing of Calcutta from Valparaiso, S. 64 c 
20' 17.4" E. ; of Valparaiso from Calcutta, S. 54° 54' 25.2" W. 

3. Latitude, 49° 58' 23.1" X 

§ 170; page 113. 

1. Time, 6 h. m. 43 s. a.m. ; longitude, 44° 49' 18" W. 

2. 15° 0' 41.4". 3. N. 56° 28' 8.5" E, 4. 5 h. 3 m. 27 s. a.m. 



3. 


a= 68° 46' 28.4", 


b= 73° 47' 57.8", 


4. 


a= 90° 53' 2.6", 


b = 117° 48' 59.6", 


5. 


a = 103° 31' 33.8", 


b= 53° 4' 26.2", 


6. 


b = 85° 48' 53.8". 


§ 165 ; page 109. 


4. 


C= 65° 29' 1.0", 


A= 97° 18' 33.8", 


5. 


B = 42° 40' 9.2", 


C=159°54' 3.6", 


or, 


B = 137° 19' 50.8", 


C= 50° 21' 16.4", 


6. 


Impossible. 




7. 


C= 90°, 


B = 113° 33' 15.5", 


8. 


B= 68° 17' 2.4", 


A = 132° 35' 12.4", 


or, 


B = 111° 42' 57.6", 


A= 77° 3' 48.0", 


9. 


Impossible. 




10. 


0=146° 37' 40.2", 


B= 55° 1'11.8", 
§ 166 ; page 111. 


2. 


b = 114° 48' 57.9", 


a= 82° 54' 0.0", 


3. 


a= 67° 25' 2.3", 


c = 160° 6' 10.0", 


or, 


a = 112° 34' 57.7", 


c = 103° 6' 20.4", 


4. 


c= 90°, 


B= 63° 46' 30.2", 


5. 


Impossible. 








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LIBRARY OF CONGRESS 



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